I'm Reviewing 10.1
we weren't allowed to use calculators and we had to know the trig chart . . .
In order to do this you must know these formulas:
cos(alpha +/- beta)= cos(alpha)cos(beta) -/+ sin(alpha)sin(beta)
sin(alpha +/- beta)= sin(alpha)cos(beta) +/- cos(alpha)sin(beta)
Alpha and beta will come from the trig chart. You will either add or subtract them to get the angle needed.
Ex: find the exact value of sin15 degrees.
alpha= 45
beta= 30
45-30 = 15
sin(45-30)= (sin 45)(cos 30) - (cos 45)(sin 30)
=(square root of 2/2)(square root of 3/2) - (square root of 2/2)(1/2)
square root of 6/4 - square root of 2/4 = square root of 6 - square root of 2/4
sin 15= square root of 6 - square root of 2/4
Thursday, December 30, 2010
Malorie's Holiday blog #3
I'm reviewing 9.1 right triangles
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
Malorie's Holiday Blog #2
I'm reviewing 8.1 because it was easy.
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
Malorie's Holiday BLog #1
In this blog i'm looking back to the easiest stuff I learned. 7-4 reference angles.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
Charlie's Christmas Holiday's Blog #4
one recent thing we did was arithmetic sequences. where you use the formula to find the certain number of the sequence that you want to find or just find the real formula to find that certain number of the sequence. the formula you use is t*n = t*1 + (n - 1) d; where t*n is the sequence anser you want to find, t*1 is the firse numbe of the sequence, and d is the number being added.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
Charlie's Christmas Holiday's Blog #3
this first semester we also did law of sines. the formula (sinA/opposite leg = sinB/opposite leg) is used when you have a nonright triangle and have atleast one pair of an angle and oposite leg with either a different leg or another angle. to work this you just plug into the formula and cross mulitply like we use to do in like prealgebra back in the gap.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
Charlie's Christmas Holiday's Blog #2
in another chapter we went through sine, cosine, and tangent for right triangles. for this we use the 'word' SOHCAHTOA. this means that sine equals opposite over hypotenuse, cosine equals adjacent oer hypotenuse, and tangent equals opposite over adjacent. to do this you are given the information that the triangle is a right triangle and either a side length and an angle other than the 90 degree one or two side lengths. in order to solve, though, a triangle must be drawn.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
Charlie's Christmas Holiday's Blog #1
we went through captial sigma one section in the first semester, where there is a big E looking thing called captial sigma, a limit at the top of it [#], a limit with an index at the bottom of it [x = #], and the summand on the right of the big E thing [f(x)]. to solve you plug the bottom limit into the summand that's on the right of the captil sigma and add it with the same formula but the next numberical number until you get to the limitation that is on the top of the big E thing.
For example:
4
E p+3
p=2
(2+3) + (3+3) + (4+3)
= 5 + 6 + 7
= 18
~~>for this you pluged the 2 into the formula p + 3, then added it with the same formula but with 3 instead of 2, then again with 4 instead of 3, after that you couldn't go any further because the limit at the top was 4. then you solve to get 18.
For example:
4
E p+3
p=2
(2+3) + (3+3) + (4+3)
= 5 + 6 + 7
= 18
~~>for this you pluged the 2 into the formula p + 3, then added it with the same formula but with 3 instead of 2, then again with 4 instead of 3, after that you couldn't go any further because the limit at the top was 4. then you solve to get 18.
Wednesday, December 29, 2010
lawrence's holiday blog 2
we did chapter 9-1. It talked about solving for right triangles and law of sines.
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
lawrence's holiday blog 1
we learned the first part of chapter 10. it is really easy. just know the formula and plug in.
formulas:
cos(alpha + or - beta) =sine alpha cos beta - or + sin alpha cos beta
sin (alpha + or - beta)= sin alpha cos beta + or - cos alpha sin beta
* aplha and beta come from the trig chart and add or subtract to get the angle you are looking for.
ex. find the exact value of sin 15
alpha= 45
beta= 30
sin(45-30)= sin45 cos30- cos45 sin 30
square root of2/2 square root of3/2 - square root of2/2 1/2
square root of6/4- square root of2/4 = square root of 6-2/f
ex.2
suppose that sin alpha=4/5 and sin beta= 5/13. where 0
cos alpha cos beta- sinalpha sin beta
(3/5)(-12/13)- (4/5)(5/13)
(-36/65)-(20/65)
(-56/65)
that is all that i learned in chapter 10 section 1. i hope this helped out the people who dont understand even though i dont know how you wont get this section when it is the easiest thing we probably will cover.
formulas:
cos(alpha + or - beta) =sine alpha cos beta - or + sin alpha cos beta
sin (alpha + or - beta)= sin alpha cos beta + or - cos alpha sin beta
* aplha and beta come from the trig chart and add or subtract to get the angle you are looking for.
ex. find the exact value of sin 15
alpha= 45
beta= 30
sin(45-30)= sin45 cos30- cos45 sin 30
square root of2/2 square root of3/2 - square root of2/2 1/2
square root of6/4- square root of2/4 = square root of 6-2/f
ex.2
suppose that sin alpha=4/5 and sin beta= 5/13. where 0
cos alpha cos beta- sinalpha sin beta
(3/5)(-12/13)- (4/5)(5/13)
(-36/65)-(20/65)
(-56/65)
that is all that i learned in chapter 10 section 1. i hope this helped out the people who dont understand even though i dont know how you wont get this section when it is the easiest thing we probably will cover.
Tuesday, December 28, 2010
Helen's Holiday Blog #4
9-1
Examples:
1) The safety instructions for a 20ft ladder indicate that the ladder should not be inclined at more than 70 degrees with the ground. Suppose the ladder is leaned against a house at this angle, as shown. Find the distance from the base of the house to the foot of the ladder and the height reached by the ladder.
cos70=a/20
a=20cos70=6.840ft
sin70=b/20
b=20sin70=18.794ft
2) The highest tower in the world is in Toronto, Canada and is 553m high. An observer at point A, 100m from the center of the tower's base, sights the top of the tower, the angle of elevation is angle A. Find the measure of this angle to the nearest tenth of degrees.
tan A=553/100
A=tan inverse (553/100)=79.750 degrees
9-2
Formulas: 1/2 bh for right angles only or 1/2 ab sin(angle b/w) for non right angles
Examples:
1) Two sides of a triangle have lengths 7cm and 4cm. The angle between the sides measures 73 degrees. Find the area of the triangle.
A=1/2ab sin (theta)
A=1/2(7)(4)sin(73)
A=1/2(28)sin73
=14 sin 73
=13.388
Examples:
1) The safety instructions for a 20ft ladder indicate that the ladder should not be inclined at more than 70 degrees with the ground. Suppose the ladder is leaned against a house at this angle, as shown. Find the distance from the base of the house to the foot of the ladder and the height reached by the ladder.
cos70=a/20
a=20cos70=6.840ft
sin70=b/20
b=20sin70=18.794ft
2) The highest tower in the world is in Toronto, Canada and is 553m high. An observer at point A, 100m from the center of the tower's base, sights the top of the tower, the angle of elevation is angle A. Find the measure of this angle to the nearest tenth of degrees.
tan A=553/100
A=tan inverse (553/100)=79.750 degrees
9-2
Formulas: 1/2 bh for right angles only or 1/2 ab sin(angle b/w) for non right angles
Examples:
1) Two sides of a triangle have lengths 7cm and 4cm. The angle between the sides measures 73 degrees. Find the area of the triangle.
A=1/2ab sin (theta)
A=1/2(7)(4)sin(73)
A=1/2(28)sin73
=14 sin 73
=13.388
Helen's Holiday Blog #3
8-5
Things you CAN'T do!!!!!!!!!!!
2.Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Examples:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = +/- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx)+(sinx-1)=0
2sinx(sinx-1)+(sinx-1)=0
(2sinx+1)(sinx-1)=0
2sinx+1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Answer: 0,71.565,180,251.565,360
- You would use the identities to get to the same trig function. If you can't try to make it tan or cot by sin/cos or cos/sin
- Follow the inverse rules
Things you CAN'T do!!!!!!!!!!!
- Divide by a trig function when solving to cancel.
2.Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Examples:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = +/- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx)+(sinx-1)=0
2sinx(sinx-1)+(sinx-1)=0
(2sinx+1)(sinx-1)=0
2sinx+1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Answer: 0,71.565,180,251.565,360
Helen's Holiday Blog #1
8-1
QIII: add 180 degrees
QIV: make negative, add 360 degrees
Examples:
1) 3cos(theta) +9=7
3cos(theta)=-2
cos(theta)=-2/3
theta=cos inverse (-2/3)
=48.1897 degrees
Answer: theta= 131.810,228.190
2) 5 sec (theta)+6=0
5 sec (theta)=-6
sec(theta)= -6/5
theta= sec inverse (6/5)
theta= cos inverse (5/6)
Answer: theta= 146.443,213.557
- To solve for theta you get the trig function by itself and then take an inverse.
- An inverse has 2 answers, find where the angle is based on trig function and if the number is positive or negative.
- Steps: Take the inverse of the postive # to find the QI angle.
QIII: add 180 degrees
QIV: make negative, add 360 degrees
Examples:
1) 3cos(theta) +9=7
3cos(theta)=-2
cos(theta)=-2/3
theta=cos inverse (-2/3)
=48.1897 degrees
Answer: theta= 131.810,228.190
2) 5 sec (theta)+6=0
5 sec (theta)=-6
sec(theta)= -6/5
theta= sec inverse (6/5)
theta= cos inverse (5/6)
Answer: theta= 146.443,213.557
Helen's Holiday Blog #2
8-3
y=A sin(Bx +/- C) +/- D
or
y=A cos( Bx +/-C) +/-D
Steps:
1) Follow previous steps for period change with 5 points.
3)move ally values up or down.
Examples:
1) y= 3cos(Ï€x + 1/2)+2
amp: 3 period: 2Ï€/Ï€=2
1. o/Ï€=0 1. 0-1/2=-1/2
2. π/2/π=1/2 2. 1/2-1/2=0
3 .Ï€/Ï€=1 3. 1-1/2=1/2
4. 3Ï€/2/Ï€=3/2 4. 3/2-1/2=1
5. 2Ï€/2/Ï€=2 5. 2-1/2=3/2
Then you would graph it.
And your answer would be in a different color
y=A sin(Bx +/- C) +/- D
or
y=A cos( Bx +/-C) +/-D
Steps:
1) Follow previous steps for period change with 5 points.
- take all 5 points and divide by B
3)move ally values up or down.
Examples:
1) y= 3cos(Ï€x + 1/2)+2
amp: 3 period: 2Ï€/Ï€=2
1. o/Ï€=0 1. 0-1/2=-1/2
2. π/2/π=1/2 2. 1/2-1/2=0
3 .Ï€/Ï€=1 3. 1-1/2=1/2
4. 3Ï€/2/Ï€=3/2 4. 3/2-1/2=1
5. 2Ï€/2/Ï€=2 5. 2-1/2=3/2
Then you would graph it.
And your answer would be in a different color
Nicala's Blog
in chapter nine section 3
law of sine
you used the law of sine with non-right triangles and only when you know the angle and opposite leg's value.
formula
sin(angle)over its opposite leg= sine(angle two) over its opposite leg
you cross multiply to solve
A engineer wants to determine the distance between point A and point C. She knows that AB=25 angle A=110 and angle B=20. Find AC and BC.
first you HAVE to draw the triangle so you can solve it otherwise it will be very hard to visual what the triangle looks like.
you need to subtract the two angles away from 180.
180-110-20=50
then you but it in the formula
sin50 over 25=sin120 over AC
ACsin50=25sin20
AC=25sin20 overs sin 50
AC=11.2
law of sine
you used the law of sine with non-right triangles and only when you know the angle and opposite leg's value.
formula
sin(angle)over its opposite leg= sine(angle two) over its opposite leg
you cross multiply to solve
A engineer wants to determine the distance between point A and point C. She knows that AB=25 angle A=110 and angle B=20. Find AC and BC.
first you HAVE to draw the triangle so you can solve it otherwise it will be very hard to visual what the triangle looks like.
you need to subtract the two angles away from 180.
180-110-20=50
then you but it in the formula
sin50 over 25=sin120 over AC
ACsin50=25sin20
AC=25sin20 overs sin 50
AC=11.2
Nicala's Blog
in chapter nine section one
its the right triangle that we did in geometry.
when doing the right triangles you must remember what all the trig functions symbolize on the triangle. the easiest way to remember is to use SOHCAHTOA.
the S stands for sine and the OH- stands for what sine is on the triangle- Opposite over Hypothenuse.
its the same for the rest of it too.
CAH- cosine=adjecent over hypothenuse
TOA-tangent=opposite over adjacent
its the right triangle that we did in geometry.
when doing the right triangles you must remember what all the trig functions symbolize on the triangle. the easiest way to remember is to use SOHCAHTOA.
the S stands for sine and the OH- stands for what sine is on the triangle- Opposite over Hypothenuse.
its the same for the rest of it too.
CAH- cosine=adjecent over hypothenuse
TOA-tangent=opposite over adjacent
For the rest of the trig function, cotangent, secant, and cosecant, they are the inverse of other three trig functions i mentioned.
Cotangent is the opposite of tangent= adjacent over opposite
Secant is the opposite of cosine= hypothenuse over adjacent
and Cosecant is the opposite of sine=hypothenuse over adjacent
Hint: the hypothenuse is always the opposite of the ninety
Example
Solve for each part of the triangle.
A=35 and a=4
B=55 and b=?
C=90 and c=5
first need to draw the triangle with this information on it.
the you need to look and see what you need
since you know that this is a right triangle you should the pythagorean theorm which is a(squared)+ b(squared)=c(squared)
4(squared)+b(squared)=5(squared)
16+b=25
b=3
Nicala's Blog
in chapter seven section two
all you have to do is remember the formulas and the hints to pass this section
formulas
s=r(theta)
k-one half(r)squaredtheta
k=onehalf (r)(s)
s=arc length
r=radius
theta=central angle
k=area
word problem hints
theta=apparent size
s=diameter
r=distance between objects
Also your theta must be in radians
Example
A sector of a circle has an arc length of 8 cm and the area of 75 cm. Find its radius and the measure of its central angle.
s=8cm
k=75 cm
r=?
theta=?
using one of the formulas for area
k=one half(r)(s)
75=(1/2)(r)(8)
75=4r
then you divide four by both sides
r=18.75cm
next you use the arclength formula
s=(r)theta
8=(18.75)(theta)
theta=2.34375 rads
all you have to do is remember the formulas and the hints to pass this section
formulas
s=r(theta)
k-one half(r)squaredtheta
k=onehalf (r)(s)
s=arc length
r=radius
theta=central angle
k=area
word problem hints
theta=apparent size
s=diameter
r=distance between objects
Also your theta must be in radians
Example
A sector of a circle has an arc length of 8 cm and the area of 75 cm. Find its radius and the measure of its central angle.
s=8cm
k=75 cm
r=?
theta=?
using one of the formulas for area
k=one half(r)(s)
75=(1/2)(r)(8)
75=4r
then you divide four by both sides
r=18.75cm
next you use the arclength formula
s=(r)theta
8=(18.75)(theta)
theta=2.34375 rads
Sunday, December 26, 2010
Nathan's blog
Nothing says Merry Christmas like doing your math blogs. The first sentence of this blog is, of course, a lie. I will probably review chapter 13 for this blog.
Chapter 13-1 involved arithmetic and geometric sequences. In an arithmetic sequence, you add or subtract in order to get to the next number. In a geometric sequence you multiply or divide to get the next number.
Ex. Is the following sequnce arithmetic, geometric, or neither.
1.) 2,4,8,16,..........Geometric because you are multiplying by 2.
2.) 15,20,25,30,.......Arithmetic because you are adding by 5.
3.) 3,7,12,18,......Neither because their is not common difference.
You will need to know these formulas in order to get the next number in the sequence.
Arithmetic- tn=t1(n-1)d
Geometric- tn=t1 x r^n-1
Find the first four terms for tn=15n+7
t1=15(1)+7=22
t2=15(2)+7=37
t3=15(3)+7=52
t4=15(4)+7=67
That's all for this blog. I hope that someone learned a thing or two about chapter 13. Happy New Year (2011)!
Chapter 13-1 involved arithmetic and geometric sequences. In an arithmetic sequence, you add or subtract in order to get to the next number. In a geometric sequence you multiply or divide to get the next number.
Ex. Is the following sequnce arithmetic, geometric, or neither.
1.) 2,4,8,16,..........Geometric because you are multiplying by 2.
2.) 15,20,25,30,.......Arithmetic because you are adding by 5.
3.) 3,7,12,18,......Neither because their is not common difference.
You will need to know these formulas in order to get the next number in the sequence.
Arithmetic- tn=t1(n-1)d
Geometric- tn=t1 x r^n-1
Find the first four terms for tn=15n+7
t1=15(1)+7=22
t2=15(2)+7=37
t3=15(3)+7=52
t4=15(4)+7=67
That's all for this blog. I hope that someone learned a thing or two about chapter 13. Happy New Year (2011)!
Sunday, December 19, 2010
Nathan's week before Christmas blog
This is the first, I repeat first, of the three blogs that B-Rob demands that we do over the holidays.
In this blog, I will review chapter 7, because that was probably the easiest chapter we did all year.
The first thing that I will go over is converting degrees to radians and radians to degrees.
Convert 57 degrees to radians.
-All you do is multiply 57*Ï€/180=57Ï€/180; This reduces down to 19Ï€/60
Convert 4Ï€ to degrees.
-All you do is multiply 4Ï€*180/Ï€=720 degrees.
Now I will go over how to get co-terminal angles.
Ex. Give a positive co-terminal angle for 55.
-Now you just add 360 to 55=415 degrees
Give a positive co-terminal angle for π/2.
-You can type this into your calculator by putting 1/2+2 and then convert it to a fraction.
You are adding 2Ï€ and the final answer is:
=5Ï€/2
That is about it for this blog. Still got blog prompts to answer, so................
In this blog, I will review chapter 7, because that was probably the easiest chapter we did all year.
The first thing that I will go over is converting degrees to radians and radians to degrees.
Convert 57 degrees to radians.
-All you do is multiply 57*Ï€/180=57Ï€/180; This reduces down to 19Ï€/60
Convert 4Ï€ to degrees.
-All you do is multiply 4Ï€*180/Ï€=720 degrees.
Now I will go over how to get co-terminal angles.
Ex. Give a positive co-terminal angle for 55.
-Now you just add 360 to 55=415 degrees
Give a positive co-terminal angle for π/2.
-You can type this into your calculator by putting 1/2+2 and then convert it to a fraction.
You are adding 2Ï€ and the final answer is:
=5Ï€/2
That is about it for this blog. Still got blog prompts to answer, so................
Monday, December 13, 2010
TAYLOR'S BLOG REVIEW!!! 3 MERRY CHRISTMAS AND HAPPY NEW YEAR!!
REVIEW TIME!!
The following is part one of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Sin for these problems as well and only Sin for part 1 of this lesson.
To find Sin remember this:
Sin Theta = opposite/ adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 13
B= 80 degrees b=16
C= 10 degrees c=?
First, draw your triangle with the information that I gave you includes.
Sin 10= ?/13 Then, write out your equation like the formula I gave you.
c= 2.257 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=2
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Sin C= 6/12 Then, write out your equation like the formula I gave you.
C= Sin^-1 (6/12) Then, you need to find the inverse.
C= 30 degrees Then, solve. Then, you need to make sure that the other possible angle cannot be used so turn the answer into a negative and add 180 degrees. You should get 150 degrees. Then, add 90 degrees to it and you get 240. Since this is over 180 degrees we cannot use it for triangles only have 180 degrees in them.
180 degrees -90 degrees -30 degrees= 60 degrees Finally, you just have to subtract 90 degrees and 30 degrees from 180 degrees and you will get 60 degrees for angle B.
So your answers will be C= 30 degrees and B= 60 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 1 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part one of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Sin for these problems as well and only Sin for part 1 of this lesson.
To find Sin remember this:
Sin Theta = opposite/ adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 13
B= 80 degrees b=16
C= 10 degrees c=?
First, draw your triangle with the information that I gave you includes.
Sin 10= ?/13 Then, write out your equation like the formula I gave you.
c= 2.257 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=2
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Sin C= 6/12 Then, write out your equation like the formula I gave you.
C= Sin^-1 (6/12) Then, you need to find the inverse.
C= 30 degrees Then, solve. Then, you need to make sure that the other possible angle cannot be used so turn the answer into a negative and add 180 degrees. You should get 150 degrees. Then, add 90 degrees to it and you get 240. Since this is over 180 degrees we cannot use it for triangles only have 180 degrees in them.
180 degrees -90 degrees -30 degrees= 60 degrees Finally, you just have to subtract 90 degrees and 30 degrees from 180 degrees and you will get 60 degrees for angle B.
So your answers will be C= 30 degrees and B= 60 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 1 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
TAYLOR'S BLOG REVIEW!!! 2
REVIEW TIME!!
The following is part two of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Cos for these problems as well and only Cos for part 2 of this lesson.
To find Cos remember this:
Cos Theta = adjacent/ hypotenuse
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= 50 degrees b=?
C= 40 degrees c=2
First, draw your triangle with the information that I gave you includes.
Cos 40= ?/7 Then, write out your equation like the formula I gave you.
b= 5.362 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=8
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Cos C= 8/12 Then, write out your equation like the formula I gave you.
C= Cos^-1 (8/12) Then, you need to find the inverse.
C= 48.190 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Cos is on the first and forth quadrant, you cannot do it for the forth quadrant is up to 360 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees - 48.190 = 41.81 degrees Finally, you just have to subtract 90 degrees and 48.190 degrees from 180 degrees and you will get 41.81 degrees for angle B.
So your answers will be C= 48.190 degrees and B= 41.81 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part two of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Cos for these problems as well and only Cos for part 2 of this lesson.
To find Cos remember this:
Cos Theta = adjacent/ hypotenuse
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= 50 degrees b=?
C= 40 degrees c=2
First, draw your triangle with the information that I gave you includes.
Cos 40= ?/7 Then, write out your equation like the formula I gave you.
b= 5.362 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=8
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Cos C= 8/12 Then, write out your equation like the formula I gave you.
C= Cos^-1 (8/12) Then, you need to find the inverse.
C= 48.190 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Cos is on the first and forth quadrant, you cannot do it for the forth quadrant is up to 360 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees - 48.190 = 41.81 degrees Finally, you just have to subtract 90 degrees and 48.190 degrees from 180 degrees and you will get 41.81 degrees for angle B.
So your answers will be C= 48.190 degrees and B= 41.81 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
TAYLOR'S BLOG REVIEW!! 1
REVIEW TIME!!!!
The following is part three of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Tan for these problems as well and only Tan for part 3 of this lesson.
To find Tan remember this:
Tan Theta = opposite/adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 20
B= 30 degrees b=50
C= 60 degrees c=?
First, draw your triangle with the information that I gave you includes.
Tan 30= 50/? Then, write out your equation like the formula I gave you.
Tan 30 x ?= 50 Then you would multiply by ? And get this as your answer.
c= 86.603 Finally, just divide by Tan 30 both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= ? degrees b=6
C= ? degrees c=8
First, draw your triangle with the information that I gave you includes.
Tan C= 8/6 Then, write out your equation like the formula I gave you.
C= Tan^-1 (8/6) Then, you need to find the inverse.
C= 53.130 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Tan is on the first and third quadrant, you cannot do it for the third quadrant is up to 270 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees -53.130 degrees= 86.97 degrees Finally, you just have to subtract 90 degrees and 53.130 degrees from 180 degrees and you will get 86.87 degrees for angle B.
So your answers will be C= 53.130 degrees and B= 86.87 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part three of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Tan for these problems as well and only Tan for part 3 of this lesson.
To find Tan remember this:
Tan Theta = opposite/adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 20
B= 30 degrees b=50
C= 60 degrees c=?
First, draw your triangle with the information that I gave you includes.
Tan 30= 50/? Then, write out your equation like the formula I gave you.
Tan 30 x ?= 50 Then you would multiply by ? And get this as your answer.
c= 86.603 Finally, just divide by Tan 30 both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= ? degrees b=6
C= ? degrees c=8
First, draw your triangle with the information that I gave you includes.
Tan C= 8/6 Then, write out your equation like the formula I gave you.
C= Tan^-1 (8/6) Then, you need to find the inverse.
C= 53.130 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Tan is on the first and third quadrant, you cannot do it for the third quadrant is up to 270 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees -53.130 degrees= 86.97 degrees Finally, you just have to subtract 90 degrees and 53.130 degrees from 180 degrees and you will get 86.87 degrees for angle B.
So your answers will be C= 53.130 degrees and B= 86.87 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Sunday, December 12, 2010
Dom's blog
Chapter 13-1
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
Example 1:
3, 7, 11, 15, 19 has a1 = 3, d = 4,
and n = 5. The explicit formula is
an = 3 + (n – 1)·4 = 4n – 1
Example 2:
These two terms are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that a12 = ( a5 )( r7 ). I can use this to solve for the value of the common ratio r:
160 = (5/4)(r7)
128 = r7
2 = r
Since a5 = ar4, then I can solve for the value of the first term a:
5/4 = a(24) = 16a
5/64 = a
well thats it
Feroz's Blog
So. tired. here is. blog. 13 -1.
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n + 3
t1 = 201(1) +3
201 + 3 = 204
t2 = 201(2) +3
402 + 3 = 405
t3 = 201(3) +3
603 + 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 + (n-1)d
tn = 3 + (n-1)(4)
tn = 3 + 4n - 4
tn = 4n-1
*insert witty comment*
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n + 3
t1 = 201(1) +3
201 + 3 = 204
t2 = 201(2) +3
402 + 3 = 405
t3 = 201(3) +3
603 + 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 + (n-1)d
tn = 3 + (n-1)(4)
tn = 3 + 4n - 4
tn = 4n-1
*insert witty comment*
Kaitlyn's bloggg
In this blogg, i am going to go over 10-1. This section was probably one of the easiest ones we have done so far this year because all you have to do is know the formulas and you will know how to solve the problems. you also have to make sure you know the trig chart.
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
Ex: find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof6/4)+(squarerootof2/4)
-sin(75)=(squarerootof 6+squarerootof 2)/4 <----answerrrrrrrr!
ex: simplifyyy cos40cos20-sin40sin20
-cos(40+20)
-cos(60)
-1/2 <----answerrr!!
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
Ex: find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof6/4)+(squarerootof2/4)
-sin(75)=(squarerootof 6+squarerootof 2)/4 <----answerrrrrrrr!
ex: simplifyyy cos40cos20-sin40sin20
-cos(40+20)
-cos(60)
-1/2 <----answerrr!!
Nathan's Blog
For this blog, I will be going over section 9-3 and 9-4, which deals with the Law of Sines and the Law of Cosines.
Law of Sines:
-Only used with non-right triangles.
-Only use when you have an angle and opposite leg value you know.
Formula:
sin(angle)/opposite leg=sin(angle 2)/opposite leg 2
*Cross multiply to solve.
Ex. A civil engineer wants to determine the distance from points A and B to an inaccessible point C. From direct measurement, the engineer knows that AB=25m, angle A=110 degrees and angle B=20 degrees. Find AC and BC.
sin50/25=sin110/BC
BC=25sin110/sin50
BC=30.667 m
sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m
Law of Cosines:
-Used when you don't have a pair in a non-right triangle.
Ex. Suppose that two sides of a triangle have lengths of 3 cm and 7 cm and that the angle between them measures 130 degrees. Find the third side of the triangle using the Law of Cosines.
p^2=3^2+7^2-2(3)(7)cos130
p=√3^2+7^2-2(3)(7)cos130
p=9.22 cm
Law of Sines:
-Only used with non-right triangles.
-Only use when you have an angle and opposite leg value you know.
Formula:
sin(angle)/opposite leg=sin(angle 2)/opposite leg 2
*Cross multiply to solve.
Ex. A civil engineer wants to determine the distance from points A and B to an inaccessible point C. From direct measurement, the engineer knows that AB=25m, angle A=110 degrees and angle B=20 degrees. Find AC and BC.
sin50/25=sin110/BC
BC=25sin110/sin50
BC=30.667 m
sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m
Law of Cosines:
-Used when you don't have a pair in a non-right triangle.
Ex. Suppose that two sides of a triangle have lengths of 3 cm and 7 cm and that the angle between them measures 130 degrees. Find the third side of the triangle using the Law of Cosines.
p^2=3^2+7^2-2(3)(7)cos130
p=√3^2+7^2-2(3)(7)cos130
p=9.22 cm
lawrence's blog
on this blog i will be going over chapter 9-1. it is one of the easiest chapters. you just have to know the formulas and you will be fine.
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Holiday Blog Prompt 3
What concept(s) from Algebra II did you most struggle with? Why? We will be revisiting many Algebra II concepts this semester, what do you plan to do differently to master them?
Holiday Blog Prompt 2
Explain the different types of polar graphs and their equations. Find sites with images of each. **Hint use google and search the images tab. Images can be pasted into blogger and include the link. You should have different sites.
Holiday Blog Prompt 1
Come up with your own trig graphing problem and walk someone through it step by step explaining each step and formula as though they have never taken Advanced Math.
Friday, December 10, 2010
Taylor's 15th blog
These are my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=9 tn=5tn-1+6
t1=5 (9)+6= 51 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 5 (51) + 6 = 261
t3= 5 (261) + 6= 1311
t4= 5 (1311) + 6 = 6561
t5= 5 (6561) + 6 = 32811
So your answers will be 1311, 6561, and 32811.
Now let's try finding the recursive definition for 10, 12, 14, 16...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 2 in between each number your answer or equation will be tn-1+2.
These are the rest of my notes on Ch. 13.2. So UNTIL MONDAY SEE YALL!!!
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=9 tn=5tn-1+6
t1=5 (9)+6= 51 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 5 (51) + 6 = 261
t3= 5 (261) + 6= 1311
t4= 5 (1311) + 6 = 6561
t5= 5 (6561) + 6 = 32811
So your answers will be 1311, 6561, and 32811.
Now let's try finding the recursive definition for 10, 12, 14, 16...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 2 in between each number your answer or equation will be tn-1+2.
These are the rest of my notes on Ch. 13.2. So UNTIL MONDAY SEE YALL!!!
Nicala's Blog
chapter 11 section 1
convert to rectangular use the following formulas
x=r cos theta
y=r sin theta
y=r sin theta
and convert to polar use the formulas
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse(three over five)=31
put it on the coordinating plane
theta=211
so the answer is squared root of 34, 31 and the negative squared root of 34,211
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse(three over five)=31
put it on the coordinating plane
theta=211
so the answer is squared root of 34, 31 and the negative squared root of 34,211
Monday, December 6, 2010
Week 7 Prompt
What have you personally accomplished in math this semester? Do you feel more confident? What will you do differently next semester?
Sunday, December 5, 2010
Feroz's ASDJEOIAHFOIADS;ADIFSJ
JUST FINISHED THAT MATH PROJECT FOR BROB AND I CAN'T STAY AWAKE IF MY LIFE DEPENDED ON IT.
Section 10-1
When everyone and their mother went over this section I chose not to, just in case I had no idea what to do a blog on, and what do ya know.
Formulas:
Sin: (alpha +/- beta) = sinAcosB +/- cosAsinB
Cos: (alpha -/+ beta) = cosAcosB +/- sinAsinB
ex. Find the exact value of sin 75
sin45cos30 + cos45sin30
= sqrt(2/2)(1/2) + sqrt(2/2)sqrt(3/2)
= sqrt(6) + sqrt(2)
----------------------
4
*yawn*
Section 10-1
When everyone and their mother went over this section I chose not to, just in case I had no idea what to do a blog on, and what do ya know.
Formulas:
Sin: (alpha +/- beta) = sinAcosB +/- cosAsinB
Cos: (alpha -/+ beta) = cosAcosB +/- sinAsinB
ex. Find the exact value of sin 75
sin45cos30 + cos45sin30
= sqrt(2/2)(1/2) + sqrt(2/2)sqrt(3/2)
= sqrt(6) + sqrt(2)
----------------------
4
*yawn*
Ch. 13-1
Sequences
arithmetic-add or subtract
geometric-multiply or divide
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the # of term in geometric sequences
ex: a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
arithmetic-add or subtract
geometric-multiply or divide
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the # of term in geometric sequences
ex: a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
What did we learn this week?
Problems we have to face,
and problems we'll overcome.
What ever choice you make, you'll have to live with it,
because what's done is done.
This week we went over things,
that some eyes have already seen.
Yet, those eyes still learned,
many many things.
Holiday time is here,
and love in all in the air.
Finals our coming soon,
and in time its going to be chaos everywhere.
But minus the hard thinking,
I'm blessed to be here.
Senor year is almost over,
so adulthood is near.
Problems we'll have to face,
and problems we'll overcome.
Just think before you react,
and that problem will be over with and done.
Enjoy
Good luck on testing everyone.
and problems we'll overcome.
What ever choice you make, you'll have to live with it,
because what's done is done.
This week we went over things,
that some eyes have already seen.
Yet, those eyes still learned,
many many things.
Holiday time is here,
and love in all in the air.
Finals our coming soon,
and in time its going to be chaos everywhere.
But minus the hard thinking,
I'm blessed to be here.
Senor year is almost over,
so adulthood is near.
Problems we'll have to face,
and problems we'll overcome.
Just think before you react,
and that problem will be over with and done.
Enjoy
Good luck on testing everyone.
Dylan's blog... :/
chp. 13-1 Sequences
-a list of numbers
arithmetic-add same # to every term
geometric-multiply by same # every term
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the *// term in geometric sequences
Ex. of sequences
a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
To find the n^m term of a sequence
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find first 3 terms for tn=4n-3
t1=4(1)-3=1
t2=4(2)-5=5
t3=4(3)-3=9
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
-a list of numbers
arithmetic-add same # to every term
geometric-multiply by same # every term
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the *// term in geometric sequences
Ex. of sequences
a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
To find the n^m term of a sequence
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find first 3 terms for tn=4n-3
t1=4(1)-3=1
t2=4(2)-5=5
t3=4(3)-3=9
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
toriss.
Tori here for the last blog before exams. I am extremely worried about this exam because i dont remember anyyyythingg from chapter 8. its gonna be really hard to pass but i hope i wont be the only one in the room taking the exam.
I'm gonna go over polar/rectangular: i wish the entire exam would be this stufff!
here are your formulas:
polar:(r, theta)r =+/-squareroot of x^ 2 + y^2::tan theta=inverse of y/x
rectangular:(x,y)x=rcosthetay=rsintheta
give the polar point for (3,4)
since the second number is not a degree, you know its in rectangular
so to convert to polar, you look at your formulas
r = squareroot of 3^2 +4^2=square root of 25 = +/-5
so r=+/-5theta=tan inverse of 4/3 =53.130degrees
so your point in polar would be (5, 53.130) and (-5, 233.130) because there are two quadrants where tan is positive
To plot, you just take your coordinate plane, go the regular for the x and for the degree, you just imagine your unit circle and go the approximate amount of degrees around. and for the other point, you just mirror it.
blah blah blahh. notes notes noteess and there ya go.
I'm gonna go over polar/rectangular: i wish the entire exam would be this stufff!
here are your formulas:
polar:(r, theta)r =+/-squareroot of x^ 2 + y^2::tan theta=inverse of y/x
rectangular:(x,y)x=rcosthetay=rsintheta
give the polar point for (3,4)
since the second number is not a degree, you know its in rectangular
so to convert to polar, you look at your formulas
r = squareroot of 3^2 +4^2=square root of 25 = +/-5
so r=+/-5theta=tan inverse of 4/3 =53.130degrees
so your point in polar would be (5, 53.130) and (-5, 233.130) because there are two quadrants where tan is positive
To plot, you just take your coordinate plane, go the regular for the x and for the degree, you just imagine your unit circle and go the approximate amount of degrees around. and for the other point, you just mirror it.
blah blah blahh. notes notes noteess and there ya go.
Kaitlynnnnnn's blogggg
In this blog, i am going to go overr what we learned in chapterr 13.
Arithmetic: a sequence of numbers in which you have to add or subtract to each number to get to the next number.
Geometric: a sequence of numbers in which you have to multiply or divide each number to get to the next number.
Ex: 4,6,8,10... find the next 2 terms.
-12,14 ---because you have to add 2 to each numberrr, this sequence is arithmatic.
Ex: 3,9,27....
-81,243 ---because you have to multiply by 3 to each number this sequence is geometric.
Limitssss!
1)if top exponent is greater than the bottom, the limit is infinity.
2)if top exponent is less than the bottom, the limit is 0.
3)if top exponent is equal to the bottom, the limit is the coefficient of the top and bottom.
Ex: 2x^2-1/2x^3
The limit is 0 because the top exponent is less than the bottom.
Arithmetic: a sequence of numbers in which you have to add or subtract to each number to get to the next number.
Geometric: a sequence of numbers in which you have to multiply or divide each number to get to the next number.
Ex: 4,6,8,10... find the next 2 terms.
-12,14 ---because you have to add 2 to each numberrr, this sequence is arithmatic.
Ex: 3,9,27....
-81,243 ---because you have to multiply by 3 to each number this sequence is geometric.
Limitssss!
1)if top exponent is greater than the bottom, the limit is infinity.
2)if top exponent is less than the bottom, the limit is 0.
3)if top exponent is equal to the bottom, the limit is the coefficient of the top and bottom.
Ex: 2x^2-1/2x^3
The limit is 0 because the top exponent is less than the bottom.
Helen's Bog
Chapter 13
Definitions:
Arithmetic Sequence- when you add or subtract to get the next term in the pattern.
Geometric Sequence- when you divide or multiply to get the next term in the pattern.
Examples:
1) Find the next four term 5,11,17,23,....
The sequence is arithmetic because you would add 6 to get the next term.
5+6=11
11+6=17
17+6=23
23+6=29
29+6=35
35+6=41
41+6=47
Answer : 29,35,41,47
2) Find the next four terms 2,6,18,54
The sequence is geometric because you would multiply by 5 to get the next four terms.
2x3=6
6x3=18
18x3=54
54x3=162
162x3=486
486x3=1458
1458x3=4374
Answer: 162,486,1458,4374
Limits
1) If the top and bottom exponent is equal , then the answer would be the coefficients.
2)If the top exponent is greater than the bottom, then the answer would be infinite.
3)If the top exponent is lesser than the bottom, then the answer would be zero.
Definitions:
Arithmetic Sequence- when you add or subtract to get the next term in the pattern.
Geometric Sequence- when you divide or multiply to get the next term in the pattern.
Examples:
1) Find the next four term 5,11,17,23,....
The sequence is arithmetic because you would add 6 to get the next term.
5+6=11
11+6=17
17+6=23
23+6=29
29+6=35
35+6=41
41+6=47
Answer : 29,35,41,47
2) Find the next four terms 2,6,18,54
The sequence is geometric because you would multiply by 5 to get the next four terms.
2x3=6
6x3=18
18x3=54
54x3=162
162x3=486
486x3=1458
1458x3=4374
Answer: 162,486,1458,4374
Limits
1) If the top and bottom exponent is equal , then the answer would be the coefficients.
2)If the top exponent is greater than the bottom, then the answer would be infinite.
3)If the top exponent is lesser than the bottom, then the answer would be zero.
Lawrence's blog
in this blog imma talk about section 9-2. this section is trying to get you to find the area of a right triangle or a non right triangle and it is two very simple formulas.
first formula
1/2 bh only use this for a right triangle
second formula
1/2 bh sin(angleb/w) only use this formula for a non right triangle.
those are the two formulas now you pretty much just plug the numbers in.
ex. two sides of a triangle have lengths 7cm & 4cm. the angle between the sides measure 73 degrees. find the area of the triangle.
you use a=1/2 ab sin theta
then you just plug in the numbers to that formula
a=1/2 (7)(4) sin73 degrees
a=14sin73 degrees
a=13.399 cm squared. (final answer)
that is what i learned in section 9-2. it is very easy as long as you know the formulas and you just plug the numbers into them. that is pretty much this whole chapter. thought i would go over this since we are in the exam review and this will be our next packet to do.
first formula
1/2 bh only use this for a right triangle
second formula
1/2 bh sin(angleb/w) only use this formula for a non right triangle.
those are the two formulas now you pretty much just plug the numbers in.
ex. two sides of a triangle have lengths 7cm & 4cm. the angle between the sides measure 73 degrees. find the area of the triangle.
you use a=1/2 ab sin theta
then you just plug in the numbers to that formula
a=1/2 (7)(4) sin73 degrees
a=14sin73 degrees
a=13.399 cm squared. (final answer)
that is what i learned in section 9-2. it is very easy as long as you know the formulas and you just plug the numbers into them. that is pretty much this whole chapter. thought i would go over this since we are in the exam review and this will be our next packet to do.
Nathan's Blog
I am just going to give a review for chapter 13 for this blog.
Section 1 dealt with sequences. An arithmetic sequence is when you add or subtract to get to the next number. A geometric sequence is when you multiply or divide to get to the next number.
Ex. 5,7,9,11.... This sequence is arithmetic because you added 2.
3,6,12,24.... This sequence is geometric because you multiplied by 2.
Now, on to limits. This was a very easy section to do.
1.) If the top exponent is equal to the bottom exponent, then you take the coefficients.
2.) If the top exponent is greater than the bottom exponent, the answer is +- infinite.
3.) If the top exponent is less than the bottom exponent, the answer is 0.
n^2/10n^3=0 (2 is less than 3)
2n^4/n^4=2 (same exponent; coefficient is 2/1)
5n^7/3n^2=Infinite (7 is greater than 2)
Now we are done with chapter 13. This chapter was easier than some of the previous ones.
Section 1 dealt with sequences. An arithmetic sequence is when you add or subtract to get to the next number. A geometric sequence is when you multiply or divide to get to the next number.
Ex. 5,7,9,11.... This sequence is arithmetic because you added 2.
3,6,12,24.... This sequence is geometric because you multiplied by 2.
Now, on to limits. This was a very easy section to do.
1.) If the top exponent is equal to the bottom exponent, then you take the coefficients.
2.) If the top exponent is greater than the bottom exponent, the answer is +- infinite.
3.) If the top exponent is less than the bottom exponent, the answer is 0.
n^2/10n^3=0 (2 is less than 3)
2n^4/n^4=2 (same exponent; coefficient is 2/1)
5n^7/3n^2=Infinite (7 is greater than 2)
Now we are done with chapter 13. This chapter was easier than some of the previous ones.
Friday, December 3, 2010
Taylor's 14th(I Think XD) Blog
This is the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 6 x 5^ n-1
t1=6 x 5^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=6 x 5^ 2-1
t3=6 x 5^ 3-1
t4=6 x 5^ 4-1
t1= 6 Then you have your answers.
t2= 30
t3= 150
t4= 750
4,8,,16,32.... find the formula for the nth term.
4,8,,16,32.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 2.
tn= t1 x r ^n-1Then write out your equation.
tn= 4 x2 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
These are the rest of my notes on Ch. 13.1. So UN TIL MONDAY SEE YALL!!!
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 6 x 5^ n-1
t1=6 x 5^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=6 x 5^ 2-1
t3=6 x 5^ 3-1
t4=6 x 5^ 4-1
t1= 6 Then you have your answers.
t2= 30
t3= 150
t4= 750
4,8,,16,32.... find the formula for the nth term.
4,8,,16,32.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 2.
tn= t1 x r ^n-1Then write out your equation.
tn= 4 x2 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
These are the rest of my notes on Ch. 13.1. So UN TIL MONDAY SEE YALL!!!
Nicala's Blog
In chapter 10 section1
we can not use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
we can not use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
Ex. Find the exact value of sin 75.
sin(45+30)=sin45cos30+cos45sin30
squared root of 2 over 2 times one half plus squared root of 2 over 2 times squared root of 3 over 2=squared root of two plus six over four
Charlie's..
This week in advanced math we did chapter 13 (i think) review.
we also started to review for our mid - term exam.
To do this we did our old chapter 7 multiple choice test, our old chapter 7 free response test, our old chapter 8 multiple choice test, and our old chapter 8 free response test. Next week we'll do the rest of our old test.
For my blog, though, i"m going to go over chapter 13 stuff.
chapter 13 had to do with sequences and such.
one was arithmetic sequences.
you use the formula:
t*n = t*1 + (n-1) d
example:
#1:
4, 12, 20, 28
t*n = 4 + (n-1) 8
= 4 + 8n -8
= -4 + 8n
#2:
3, 7, 11, 15, 19
t*n = 3 + (n-1) 4
= 3 + 4n-4
= -1 + 4n
we also started to review for our mid - term exam.
To do this we did our old chapter 7 multiple choice test, our old chapter 7 free response test, our old chapter 8 multiple choice test, and our old chapter 8 free response test. Next week we'll do the rest of our old test.
For my blog, though, i"m going to go over chapter 13 stuff.
chapter 13 had to do with sequences and such.
one was arithmetic sequences.
you use the formula:
t*n = t*1 + (n-1) d
example:
#1:
4, 12, 20, 28
t*n = 4 + (n-1) 8
= 4 + 8n -8
= -4 + 8n
#2:
3, 7, 11, 15, 19
t*n = 3 + (n-1) 4
= 3 + 4n-4
= -1 + 4n
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