Dom Dom's blog

A little bit of chapter 10

Section 1:
Finding exact values of si and cos and other stuff. I understood this pretty well, just remember the formulas and the trig chart and you'll be fine.
ex. Find the exact value of sin 75
=sin (45 + 30) = sin45cos30 + cos45 sin30
= (1/2)(sqrt2/2) + (sqrt3/2)(sqrt2/2)
= (sqrt6 + sqrt2)/4
Section 2:
Same thing as section 1, but with tan and cot. Again, formulas and trig chart.
ex. Find the exact value of tan 15
=tan (45 - 15) = tan45 - tan15/1 + tan45 tan 15
= (sqrt3/3) - 1/1 + (sqrt3/3)(1)
= (sqrt3) -3/(sqrt3) + 3
= 2 - (sqrt3)
Section 3:
Double and Half-Angle formulas. Like, sin2A and cos2A and all that other nonsense.
ex. 2sin45 cos45
Sin2A formula = 2sinAcosA
So, sin2A = sin2(45)
= sin 90 = 1

mary fairy

I will be reviewing part of what we learned in chapter 10.

section something something....

its all about sin and cos
sin(alpha +/- beta) = sinalphacosbeta+/-cosalphasinbeta
cos(alpha +/-beta) = cosalphacosbeta -/+sinalphasinbeta

they could as you to find the exact value of cos 75
and B-rob told us that you use the cos formula above and use two trig chart angles that add or subtract to give you the anlge they want.
so you know that 45 and 30 are both trig chart angles and they ADD to give you 75, so you go to your cos formula and plug in the first part for addition, then you can look further down the formula and see that if its addtion in the front, that it'll be subtraction in the back.
you plug it in and get
cos45cos30-sin45sin30
so you use your brain the remeber the trig chart and you know this
square root of two over 2 * 1/2 - square root of 3 over 2 * square root of 2 over 2which simplifies to square root of 2 + square root of 6 all over 4

but if they give you something like this
sin15cos15-cos15sin15
you can know from the formula and from the sign inbetween that you have to ADD these two angles, which is the other side of the formula.
so you get sin 30
which is 1/2
yay?

Charlie

we just took test again this week. two of the things on those so called test were the formulas sin(alpha +\- beta) = sinalpha cosbeta +\- cosalpha sinbeta and cos (alpha +/- beta) = cosalpha cosbeta -/+ sinalpha sinbeta. For these you just basically plug in the numbers or letters. You make get the front or the back of the formula. Or maybe even just one angle. For example: If they give you to solve for sin (x -z). You plug the x and z into the other half of the formula so that you get sinx cosz - cosx sinz. This doesn't simplify any smaller. Another example: if they give you the problem cosl cosp + sinl sinp. you have to simplify into either sin(alpha +/- beta) or sin (alpha +/- beta). In this formula you recognize that it is the cos formula. So to simplify you take the alpha and the beta and notice the middle sign is +. So the answer is cos (l - p)

Feroz's Blog

I'll be going over a little bit of all the sections of chapter 10, as a review for myself. Yeah.

Section 1:

Finding exact values of sin and cos and other stuff. I understood this pretty well, just remember the formulas and the trig chart and you'll be fine.

ex. Find the exact value of sin 75

=sin (45 + 30) = sin45cos30 + cos45sin30

= (1/2)(sqrt2/2) + (sqrt3/2)(sqrt2/2)

= (sqrt6 + sqrt2)/4

Section 2:

Same thing as section 1, but with tan and cot. Again, formulas and trig chart.

ex. Find the exact value of tan 15

=tan (45 - 15) = tan45 - tan15/1 + tan45tan 15

= (sqrt3/3) - 1/1 + (sqrt3/3)(1)

= (sqrt3) -3/(sqrt3) + 3

= 2 - (sqrt3)

Section 3:

Double and Half-Angle formulas. Like, sin2A and cos2A and all that other nonsense.

ex. 2sin45cos45

Sin2A formula = 2sinAcosA

So, sin2A = sin2(45)

= sin 90 = 1

That's all. Wasn't too hard, easier than Chapter 8 and maybe 9, but that's because I didn't study those. Oh well.

Helen's Blog

Section 10-1

Formulas:

cos(alpha +/- beta) cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) sin alpha cos beta +/- cos alpha sin beta

Examples:

Ex1:Find the exact value of sin15

sin(45-30)=sin45cos30-cos45sin30
=(√2/2)(√3/2)-(√2/2)(1/2)
=√6-√2/4


Section 10-2 ( same concept but tan instead)

Formula:

tan(alpha + beta)= tan (alpha) + tan (beta) / 1-tan(alpha) tan(beta)

tan(alpha - beta)= tan(alpha)- tan (beta) / 1+tan(alpha)tan(beta)

Ex2:

Simplify: tan27+tan18/1-tan27tan18

=tan(27+18)
=tan45
=1

tori's

Doing these blog thingys is really getting old. but at least im remembering to do them now. This week was a tough week because it felt like all the classes were piling things on us and then we had two tests, free response at that. but at least we get points for just putting down a formula. i'm just gonna go over section 10-1 because that was the easiest one and i know i aced all of those on the testt!



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 come from the trig chart and add or subtract to get the angle you are looking for.

ex.1 find the exact value of sin 15
alpha= 45 beta=30 sin(45-30)
you plug those into your formula then you should get::
square root2/2 ( square root3/2)- square root2/2(1/2)
square root6/4- square root2/4
=square root6- square root2/ 4



sincerely, Tori Triche :)

lawrence's blog :.(

last week was pretty rough for me. I had alot of tests and on top of that i was worrying about our adv. test. i was doing anything and everything i could do to know the formulas by heart. the first test i failed, but i know the second test i passed. hopefully chapter 11 will be easy but i doubt it. nothing in adv. ath can be easy for us.

i will go over chapter 10 section 1 again because that eas the easiest part and i wish every section was like this one.

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 come from the trig chart and add or subtract to get the abgle you are looking for.

ex.1 find the exact value od sin 15
alpha= 45
beta=30

sin(45-30) you plug those into you formula then you should get

square root2/2( square root3/20)- square root2/2(1/2)

square root6/4- square root2/4
=square root6- square root2/ 4

Kaitlyn's Bloggggggg

In this blog, i will explain a little bit about each section we learned so far in chapter 10.

10-1
This section involves finding the exact values of a trigonomic function

Find the exact value of sin15
=sin(45-30)=sin45cos30-cos45sin30
=(squarerootof 2/2)(squarerootof 3/2)-(squarerootof 2/2)(1/2)
=squarerootof 6-squarerootof 2/4


10-2
This section is pretty much the same as 10-1, but it involves tan instead of sin and cos.

Simplify tan50+tan10/1-tan50tan10
=tan(50+10)
=tan60
=squarerootof 3


10-3
This section is kinda like chapter 8, but it has more formulas and is more complex.

Simplify 2sin30cos30
=sin2(30)
=sin60
=squarerootof 3/2


Chapter 10 is pretty easy for me, if you know all your formulas and your trig chart you can easily pass this chapter..even without a calculator(:

Chapter 10 ( Rico Allen)

This week was long,
and these formulas were hard to remember.
I thought it was going to get easier,
because we're so close to December.

Santa Clause is nowhere near,
so the presents we want, we will not receive.
Just new formulas and hard work,
to keep our minds from ease.

Chapter 10 was difficult,
But I know there are many chapters to come.
I thought senor year was going to be problem free,
but my problems have just begun.

Everything happens for a reason,
and my God works for the good.
I wouldn't drop this class to ease my mind.
even if I could.

This will prepare me for life,
because like math problems, life problem we'll have to face.
To succeed, you need formulas and a good work ethic
Just to win this "problem" race.

I also learned, he puts people in your life,
to help you to succeed.
because of Mary and Ms. Robinson
this statement, I truly believe.

They have helped me get through
this whole time, while I sat in this new place.
Now its about to be another week of learning,
its time to sit down and buckle up for this new race!

Much knowledge I will receive
and many problems I'll overcome.
If learning a lot in a little amount of time was a game,
I have already won. :)

Sincerely,
Rico Allen

Thank you Ms. Robinson & Mary (if that's how you spell it) for all that you have done.

Formulas for the week:
10.1
cos(x+/- B) = cosx cosB -+ sinx sinB
sin(x+/-B) = sinx cosB +- cosx sinB

10.2
tan(x+B)= tanx+tanB/1-tanxtanB
tan(x+B)= tanx-tanB/1+tanxtan

cot(x+B)= 1-cotxcotB/cotx +cotB
cot(x-B)= 1+cotxcotB/cotx-cotB
10.3

sin 2x=2sinxcosx
cos 2x= cos^2x- sin^2x

Example:
10.1
sin 15
(45-30) =sin45cos30-cos45sin30
then solve using trig chart

I would put more examples but I have to go workout.

Enjoy!

Nathan's Blog

For this blog, I will go over a few examples of each section. This week, we basically prepared for the two chapter ten free response tests.

We couldn't use calculators in this whole chapter, and no one liked that.

Section 10-1
Find the exact value of sin15
sin(45-30)=sin45cos30-cos45sin30
=(√2/2)(√3/2)-(√2/2)(1/2)
=√6-√2/4

cos50cos10-sin50sin10
=cos(50+10)
=cos(60)
=1/2

sin5π/12cosπ/12+cos5π/12sinπ/12
=sin(5π/12+π/12)
=sin6/12
=sinπ/2
=1

Section 10-2
This is basically section 10-1 but it involves tan instead of sin and cos.

tan27+tan18/1-tan27tan18
=tan(27+18)
=tan45
=1

tan(x+π/4)=tanx+tanπ/4/1-tanxtanπ/4
=tanx+1/1-tanx

Section 10-3
In this section we used a little bit of chapter 8, in how we simplified the problem, but we used formulas from chapter 10.

2sin10cos10=sin2(10)=sin20
1-2sin^2 35=cos2(35)=cos70
1-sin^2x=cos^2 x
2sin3(alpha)cos3(alpha)=sin2(3 alpha)=sin6(alpha)

Overall, I thought that chapter ten was a semi-hard, semi-easy chapter. There were a ton of formulas that you needed to know. If you knew them, then the chapter was pretty easy,but if you failed to study formulas, then it was a very tough chapter.

Nicala's Blog

in chapter ten section two
all you have to is plug the problems into the formula. if your not using a formula then you are doing the problem WRONG!!

the formulas

tan(alpha plus beta)= tan (alpha) + tan (beta) over 1-tan(alpha) tan(beta)

tan(alpha minus beta)= tan(alpha)- tan (beta) over 1+tan(alpha)tan(beta)


example

What is the exact value of tan(15º)? We can use a difference angle formula noticing that 15º = 45º - 30º.

tan(15º)

= tan(45º - 30º)

(tan45-tan30)= tan45-tan30
1+tan45tan30

tan(15)=(square root of 2)- one half
1+(square root of 2)(one half)

tan(15)=squared root 2-3

Week 2 Blog Prompt

How do you solve a quadratic trig equation? Give two examples.

Dylans

We learned some new formulas... yay!

Formulas:
tan( A+B)= tanA+ tan B/1-tanA tanB
tan( A-B)= tanA- tan B/1+tanA tanB
Fore cot, you just flip the tan of it:
cot(A+B)= 1-cotA
cot B/ cot A+ cot B

They are pretty easy to remember.

Ex.

tan a= 1/3 tan b=1/2 find tan (a+b)

tan a+ tan b/ 1- tan a tan b
(1/3 + 1/2) /(1-(1/3 )(1/2))
(2/6 + 3/6)/(1-1/6)
(5/6)/(5/6)=1

Dom

Chapter 10-2

Tangent/Cotangent formulas:

tan( A+B)= tanA+ tan B/1-tanA tanB
tan( A-B)= tanA- tan B/1+tanA tanB
cot is reciprocal :
cot(A+B)= 1-cotA
cot B/ cot A+ cot B

Ex: Simplify tan40+tan20/1+tan40tan20

tan(40+20)
tan(60)
square root of 3

Ex 2:tan(27) + tan(18)/1 - tan(27)tan(18)

tan(a + b) = tan (27 + 18) = 45
tan(45) = 1

Well thats it

Tori's.

I am going to go over section 10-2. its almost exact to section 10-1, but this section uses tangent and cotangent.


Formulas:
tan( A+B)= tanA+ tan B/1-tanA tanB
tan( A-B)= tanA- tan B/1+tanA tanB
cot is reciprocal :
cot(A+B)= 1-cotA
cot B/ cot A+ cot B

Examples:

Ex1: suppose tan a= 1/3 tan b=1/2 .
find tan (A+B).
tan A+ tan B/ 1- tan A tan B=
(1/3 + 1/2) /(1-(1/3 )(1/2))=
(2/6 + 3/6)/(1-1/6)=(5/6)/(5/6)=1

Ex2: Simplify tan 110 degrees - tan 50 degrees/ 1+(tan 110 degrees)(tan 50 degrees)=
tan (110-50)=tan 60=
sq. root 3

In this blog i will be explaining the new sections we learned in chapter 10 this week. We went over section 10.1, 10.2, and 10.3.

Some of the formulas in section 10.1:

cos(x+/- B) = cosx cosB -+ sinx sinB
sin(x+/-B) = sinx cosB +- cosx sinB

Some of the formulas for 10.2:

tan(x+B)= tanx+tanB/1-tanxtanB
tan(x+B)= tanx-tanB/1+tanxtan
cot(x+B)= 1-cotxcotB/cotx +cotB
cot(x-B)= 1+cotxcotB/cotx-cotB

Sme of the formulas in 10.3;

sin 2x=2sinxcosx
cos 2x= cos^2x- sin^2x

Ex.1

simplify:

tan110-tan50/1+tan110tan50
tan(110-50)
tan(60)
=(Square root)3

Ex.2tan140-tan50/1+tan140tan50

tan(140-50)
tan(90)
=undefined

Helen Melon!!!!!

I'll be going over section 10-2.
In this section it kinda the same thing as section 10.1 but it goes over tan instead.

Formulas:

tan( A+B)= tanA+ tan B/1-tanA tanB

tan( A-B)= tanA- tan B/1+tanA tanB

cot is reciprocal : cot(A+B)= 1-cotA cot B/ cot A+ cot B


Examples:

Ex1: suppose tan a= 1/3 tan b=1/2 . find tan (A+B).

tan A+ tan B/ 1- tan A tan B
=(1/3 + 1/2) /(1-(1/3 )(1/2))
=(2/6 + 3/6)/(1-1/6)
=(5/6)/(5/6)
=1

Ex2: Simplify tan 110 degrees - tan 50 degrees/ 1+(tan 110 degrees)(tan 50 degrees)
=tan (110-50)
=tan 60
= sq. root 3

Feroz's Blog

I was absent the day we went over 10-2, so I'm going to attempt to learn it via this blog post. Hopefully.

According to everyone else, it's the same as 10-1 but instead of cos and sin, it deals with tan and cot. The formulas are:

tan(a + b) = tan(a) + tan(b)/1 - tan(a)tan(b)
tan(a - b) = tan(a) - tan(b)/1 + tan(a)tan(b)

cot(a+b) = 1 - cot(a)cot(b)/cot(a) + cot(b)

So then, an example:

tan(27) + tan(18)/1 - tan(27)tan(18)

Using this, we know we have to use the first formula I posted, since it matches the problem.

So tan(a + b) = tan (27 + 18) = 45

And tan(45) = 1

There you go. This seems pretty simple, though I'm sure I'm missing something.

Kaitlyn's Bloggggggg!

In this blog, i will be going over 10-2. This is just like 10-1, but it involves tan instead of sin and cos.


Formulas: 1) tan(A+B)=tanA+tanB/1-tanAtanB
2) tan(A-B)=tanA-tanB/1+tanAtanB



cot is the reciprocal of tan: cot(A+B)=1-cotAcotB/cotA+cotB



Examplessss:


Find tan15
=tan(60-45)=tan60-tan45/1+tan60tan45
=(squarerootof 3)-(1)/1+(squarerootof 3)(1)
= -1



Simplify tan40+tan20/1+tan40tan20
=tan(40+20)
=tan(60)
=squarerootof 3

Thiss is whatt we learnedd in 10-2. I likeee this chapterr, it's prettyy easy because all you have to do is plug into your formula!!!! (:

MaRy

For this blog, i will be explaining section 10.2, i accidently deleted my blog, so now i must write it again and i'm not to happy about it...

You can't do anything unless you know these formulas:

tan(a+b)=tan(a)+tan(b)/1-tan(a)tan(b)
tan(a-b)=tan(a)-tan(b)/1+tan(a)tan(b)

they could tell you something like "suppose tan(.." and you will find an answer by using the full formula, if they tell you to simplify, you will use the opposite side of the equation.

example:

simplify

tan27+tan18/1-tan27tan18

so you know you need to use the other side of the formula

which is tan(a +b)

which is tan (45) which = 1

yay

math is fun...

example 2: tan( x + pie/4)

so you find the formula that matches

so you plug your numbers in.

so you get

tanx+tanpie/4/1-tanxtanpie/4

which simplifies to,

tanx+1/1-tanx

or they could tell you this

"suppose tan alpha=1/3 tan alpha beta=1/2

and then you'll put that straight into the formula and you get:

1/3 + 3/6/1-1/6=1

but if it had said cos alpha = 1/3, you would have to draw out your triangle and take the tan of it.

yeaa!

Charlie's.

In Chapter 10 we learned formulas and how to solve by plugging into those formulas.
In this blog, though, i'm just going to do sections 10-1 and 10-2.

the Formulas for these two sections are:
sin(alpha +\- beta) = sin(alpha) cos(beta) +\- cos(alpha) sin(beta)
cos(alpha +\- beta) = cos(alpha) cos(beta) -\+ sin(alpha) sin(beta)
tan(alpha +\- beta) = tan(alpha) +\- tan(beta) / 1 -\+ tan(alpha) tan(beta)
cos(alpha +\- beta) = 1 -\+ cot(alpha) cot(beta) / cot(alpha) +\- cot(beta)

One thing is you'd get the answer of the second half of the formula and find the begining.

For example:
sin20 cos60 + cos20 sin60
to simplify you take the alpha and beta and the symbol inbetween the two multiplication things.
so... sin20 cos60 + cos20 sin60 = sin(20 + 60) = sin80

Another thing is you'd get the beginning of the formula and have to solve for the end.

For example:
cos(60-32)
to solve you just expand into the formula.
so... cos(60-32) = cos60 cos32 + sin60 sin32

Rico Allen What I learned this week.

Another long week,
but it was worth the struggle.
I learned more about Sin and Cosine,
and chapter 10.3 gave me some trouble.
I beleieve I passed my test, thats
the results of of studying hard.
But this next section,
Im going to have to take charge.
Study harder, and sleep a great deal,
are my plans.
Some things you have to sacrafise,
just to have a chance.
I also learned a life lesson,
while sitting in 6th period, in my seat.
Its not to good to complain,
it shows that you're a little weak.
When your faced with adversity,
you have to learn how to get throught it.
He'll never put you through something,
if he knew you couldn't do it.
So my goal for this weeek coming,
is try harder and get through the work witch seems unstoppable.
Im know I can do it,
because through him, anything is possible.

This week we learned some new formulas for chapter 10:
10.1
cos(x+/- B) = cosx cosB -+ sinx sinB
sin(x+/-B) = sinx cosB +- cosx sinB

10.2
tan(x+B)= tanx+tanB/1-tanxtanB
tan(x+B)= tanx-tanB/1+tanxtan

cot(x+B)= 1-cotxcotB/cotx +cotB
cot(x-B)= 1+cotxcotB/cotx-cotB
10.3

sin 2x=2sinxcosx
cos 2x= cos^2x- sin^2x

Lawrence's blog :p

i will go over section 10 -2 . 10-2 and 10-1 are exactly the same except one involves sin and cos. the other which is 10-2 goes over tan. you have to know these formulas so you can work the problems. if you dont know the formulas you will be lost and wont be able to work the problem. there arent many formulas for these two sections so just know them so you can pass and make b-robs life easier with grading our quizzes and tests.

formulas for 10-2

tan(a+b)= tan(a)+tan(b)/1- tan(a)tan(b)

tan(a-b)= tan(a)-tan(b)/ 1+tan(a)tan(b)

if they give you cot, just flip those two problems and there is your cot. that is really amazing isnt it!


ex. suppose tan(a)=1/3 tan(b)= 1/2. find tan(a+b)

=1/3+1/2
1-1/3(1/2)

= 2/6+3/6
1-1/6

= 5/6
5/6

=1



ex. tan110- tan50
1+tan110 tan50

=tan(110-50)

=tan 60

= square root of 3

Nathan's Blog

In this blog, I will go over section 10-2 which is basically 10-1, but it involves tan instead of sin and cos. It's just two more formulas that we have to memorize, but these are really easy to remember and apply.

tan(a+b)=tan(a)+tan(b)/1-tan(a)tan(b)
tan(a-b)=tan(a)-tan(b)/1+tan(a)tan(b)

cot is the reciprocal
cot(a+b)=1-cot(a)cot(b)/cot(a)+cot(b)

tan(x+π/4)
=tan(x)+tan(π/4)/1-tan(x)tan(π/4)
=tan(x)+1/1-tan(x)

Simplify

tan110-tan50/1+tan110tan50
=tan(110-50)
=tan(60)
=√3

tan27+tan18/1-tan27tan18
=tan(27+18)
=tan(45)
=1

tan140-tan50/1+tan140tan50
=tan(140-50)
=tan(90)
=undefined

Another type of problem you will see is one like this:

Suppose tan(a)=1/3 and tan(b)=1/2, find tan(a+b).
=(1/3)+(1/2)/1-(1/3)(1/2)
=(2/6)+(3/6)/1-(1/6)
=(5/6)/(5/6)
=1

Now, I will go to section 10-3, where we reviewed how to simplify equations using techniques from chapter 10 and chapter 8. Also, in this section, we learned about ten new formulas, which all come into play at some time.

1.)2sin10cos10=sin2(10)=sin20
2.)cos^2 15-sin^2 15=cos2(15)=cos30=√3/2
3.)1-sin^2 35=cos2(35)=cos70
4.)2cos^2 25-1=cos2(25)=cos50
5.)2tan50/1-tan^2 50=tan2(50)=tan100
6.)1-sin^2 x=cos^2 x
7.)2sin3(a)cos3(a)=sin2(3a)=sin6(a)
8.)cos^2 5y-sin^2 5y=cos2(5y)=cos10y

R.I.P. Bradly Gaude, you are greatly missed down here by your family, friends, and all of Riverside.

Malpal's Blog

10.2

in this lesson we did some stuff with tan.

to do this you must know these formulas:
tan(alpha + beta)=tan alpha + tan beta/1- tan alpha x tan beta
tan(alpha - beta)=tan alpha - tan beta/1- tan alpha x tan beta

cot is reciprocal
cot(alpha + beta)= 1- cot alpha x cot beta/ cot alpha + cot beta

Example:

Simplify:
tan 110 - tan 50/ 1 + tan 110 x tan 50

use the formula to figure out what you need to do

=tan(110-50)
=tan(60) this is on the trig chart so it can be simplified
= square root of 3

Taylor's 8th(or 9th I don't know XD) Blog

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= 60
B= 40 degrees b=40
C= 50 degrees c=?

First, draw your triangle with the information that I gave you includes.

Tan 40= 40/? Then, write out your equation like the formula I gave you.

Tan 40 x ?= 40 Then you would multiply by ? And get this as your answer.

c= 47.670 Finally, just divide by Tan 40 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= 5
B= ? degrees b=4
C= ? degrees c=2

First, draw your triangle with the information that I gave you includes.

Tan C= 2/4 Then, write out your equation like the formula I gave you.

C= Tan^-1 (2/4) Then, you need to find the inverse.

C= 26.565 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 -26.565 degrees= 63.435 degrees Finally, you just have to subtract 90 degrees and 26.565 degrees from 180 degrees and you will get 63.435 degrees for angle B.

So your answers will be C= 26.565 degrees and B= 63.435 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)

Nicala's Blog

chapter ten section one
in chapter we are not using calculators so if you do not know your trig chart you will fail!

For section one you only have memorize two formulas.

first formula: cosine (alpha plus or minus beta)= cosine alpha cosine beta minus or plus sine alpha sine beta

second formula:sine (alpha plus or minus beta)= sine alpha cosine beta minus or plus cosine alpha sine beta

example

find the exact value of cosine 75 degrees

alpha= 45
beta= 30

cosine(45+30)= cosine 45 cosine 30- sine45 sine30

cosine(75) =(square root of 2 divided by 2)(square root 3 divided by 2)-(square root of 2 divided by 2)(one divided by two)

Do not simplify the problem because need to have the same denominatior

cosine(75)=(square root of six divided by four)(square root of two divided by four)

and you leave as that for your answer

Week 8 Blog Prompt

Why is it important to know the sum/difference & Double/Half angle identities? Also, if you use a citation it needs to be different for each person. Too many people are copying answers from people and posting them.

David

section 10.1 with NO CALCULATORS

everybody else explained the hard part, so i'll go ahead and explain the easier part.
formulas: cos(α+/-β) = cos(α)cos(β) -/+ sin(α)sin(β)
sin(α+/-β) = sin(α)cos(β )+/- cos(α)sin(β)

You change the sign for cos.

Example:
they will give a satanic equation such as
sin75cos15+cos75sin15
and they as you to simplify
you look at the type of formula, and you can tell that you need the sin one
you use the numbers and put them in on the other side of the equation
so you would put sin(75+15) ( alpha + beta) and that equals sin90, which is pie/2 which is also 1, i'm not sure which on to put but yea. and its that easy

Example 2:
its the same process except they give you the other side of the equation

cos( pie + x) = -cosx
and you have to prove
you just take the first part of the equation and put it in the cos equation
pie is alpha and x is beta
after you complete this , if you've done it right, you will be proving the answer they have already given you
amen.

Dom's blog

Chapter 10-1 in this lesson, we are not allowed to use calculators.
also you have to know the trig chart in order to do this lesson.
you must know this.

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 30
Find the values of alpha and beta from the trig chart that add or subtract to get 30. Which would be 60 and 30.

sin(60-30) = (√3/2)(√3/2) - (1/2)(1/2)

sin(30) = 3/4 - 1/4

sin(30) = 1/2

And thats about it for this lesson.

Maryy asdokvnoerinakjdlaksdf k?

This friday, we did some learning, and i'm gonna show you.

Section 10.1

This whole section is about finding the sin and cos of certain angles that aren't already on the trig chart. and oh, no calculators

Here's your formulas :

Cos( alpha +/- beta) = cosalphacosbeta-/+sinalphasinbeta
Sin(alpha+/- beta) = sinalphacosbeta+/-cosalphacosbeta
(-note the plus and minuses*)

Example!
sin 15
ok, the process is that you find two numbers on the trig chart that add or subtract to get the number, and use them as alpha and beta. in this case, you can use two different combinations but we'll go with 45 and 30 so alpha =45 and beta =30
plug in your formula which you know is the sin version because of the trig function.
you get sin(45-30) =sin45cos30-cos45sin30
you take the values from the trig chart and subtract and you get
square root of 6 - square root of 2 over 4.
this section is really simple, i'm hoping it stays simple.

Feroy's Blog #?

I can't think of anything cleaver to say at the moment, so I'll just start the blog.

I'm jumping on the Chapter 10 bandwagon, so yeah, don't expect anything original.

The best part about this chapter is that I don't have to use a calculator, which is awesome, because I hate them. So does Annie and Alex Tir, so it must be an Asian thing.

Anyway, two formulas to know:

cos(α+/-β) = cos(α)cos(β) -/+ sin(α)sin(β)
sin(α+/-β) = sin(α)cos(β )+/- cos(α)sin(β)

Ex. Find the exact value of sin 30

Find the values of alpha and beta from the trig chart that add or subtract to get 30. Which would be 60 and 30.

* I'm not 100% sure on this, but alpha is always the bigger number. Again, I'm not 100% on this.

Since we're finding the exact value of sin 30, I want you to guess what formula I'm gonna use. Go ahead, there are no wrong answers, except cos.

sin(60-30) = (√3/2)(√3/2) - (1/2)(1/2)

sin(30) = 3/4 - 1/4

sin(30) = 1/2

That's it.

I actually know what I'm doing so far, and I'm fairly confident I'm going to do better this nine weeks. Well, I hope so.

dylan's 10.1 blogs :D

YAY!

10.1

In 10.1 we learned how to find exact value without calculators. All of the problems are either simplifying or using the trig chart to simplify.

formulas:

cos(alpha+/-beta)=cos(alpha)cos(beta)-/+sin(alpha)sin(beta)
sin(alpha+/-beta)=sin(alpha)cos(beta)+/-cos(alpha)sin(beta)

example:

find exact value of sin 15 degrees
(use two degree values from the trig chart to add or subtract to give you 15, so 45-30=15
alpha=45 beta=30
we will use the sin formula because the problem is sin 15 degrees
sin(45-30)=sin(45)cos(30)-cos(45)sin(30)
sin 15=(√2/2)(√3/2)-(√2/2)(1/2)
=cos 75 = (√6-√2)/4

remember, alpha and beta come from the trig chart.

kmart(: Blogggg

In this blog, i will be going over Chapter 10-1. We can't use calculators in this chapterrrr, yuckk! It was pretty easy though, i got the hang of it pretty quickly.


Formulas:
1) cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
2) sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)



Ex: Find the exact value of sin30
First you need to find the values of alpha and beta that either add or subtract to get 30

alpha: 60
beta: 30

sin(60-30)=sin60cos30-cos60-sin30

sin30=(squarerootof 3/2)(squarerootof 3/2)-(1/2)(1/2)
sin30= (3/4)-(1/4)
sin30=1/2 <------ANSWER!

I kind of likedd this section, it was prettyy basic and it was sort of easier to not use a calculator.

malorie's blog

10.1

in this lesson, we are not allowed to use calculators.
we must also know the trig chart in order to do this lesson.
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 come from the trig chart. you will either add or subtract them to get the angle needed.

Ex: Find the exact calue of sin15 degrees.

alpha=45
beta=30
45-30=15

sin(45-30)=sin45cos30-cos45sin30
=(square root of 2/2)(square root of 3/2)-cos(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

Tori's blog.

This week we went over out study guides for the exam, took the exam on Wednesday, and went over Chapter 10-Section 1 on Friday.
10.1 is a pretty easy section. the only difficult part is applying the trig chart. Since we cannot use calculators, we are relying completely on our minds and the trig chart.

The formulas we need to know are:
cos(alpha)+-(beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha)+-(beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
It is a long formula but it is easy to work out.

EXAMPLE #1
Find the exact value of sin15
to find this, you have to find two trig chart angles that add or subtract to give you 15.
(alpha)45-(beta)30=15
now, plug it into the sin formula
sin(45-30)=sin45cos30-cos45sin30
(sq.root2/2)(sq.root3/2)-(sq.root2/2)(1/2)
=(sq.root6/2)-(sq.root2/4)

This week we basically went over our study guides for our exams and took the exam. Some other stuff we did was start a new chapter 10 section 1. Chapter 10 we are not allowed to use our calculators. So for this chapter we wil have to know all the formulas. Some other things you might want to know is the trig chart. If you know to trig chart you will probably do pretty good.

Here are some of the formulas we have to know:

-cos(alpha+/- beta) = cos(alpha) cos(beta) -/+ sin(alpha) cos(beta)
-sin(alpha +/- beta) = sin(alpha) cos(beta) +/- cos(alpha) sin(beta)

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/

These are some of the things that we covered this week and the new chapter we started.

Helen Melon :)

This week we learned section 10-1.

Since you can't use your calculator, you would fail this section if you don't know your trig chart.

Formulas:

cos(alpha +/- beta)=cos alpha cos beta -/+ sin alpha sin beta

sin(alpha +/- beta)=sin alpha cos beta +/- cos alpha sin beta


*To find the angle you are looking for you would add or subtract.

Examples 1: Find exact value of Cos 15 degrees.

alpha=45 beta=30

Cos(45-30)= cos 45 cos 30+sin 45 sin 30
=(sq. root 2/2)(sq.root 3/2) + (sq. root 2/2)(1/2)
ANS= sq.root 6 + sq.root 2/4

Examples 2: Simplify:

sin 20 degrees cos 15 degrees-cos 20 degrees sin 15 degrees
=sin(20-15)
ANS= sin 5 degrees.

Lawrence"s blog (this is getting old)

i am pretty sure i can say everyone is glad that exams are over especially the one for advanced math. it was pretty hard especially since she put the one part i didnt know at all on there, but anyway 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.

What have I learned this week? (Rico Allen)

This week I learned a great deal
so knowedge I surely recieved.
At first I thought it was impossible
but soon I started to believe.

I saw that math could be simple
if you use your formulas as your tool kit.
Anything is possible only
if you put your mind to it.

I learned that math is sort of like life
and problems, you have to figure them out.
But when your done with the equation
you'll be happy without a doubt.

I learned how to write the trig chart
and it took some hard thinking.
and the difference between Sin and Cosine
was the knowledge I was seeking.

I also learned what alot of new formulas
something I thought I wouldnt be able to achieve.
like using Y over R
and converting radiants into degrees.

On friday we got a new formula
from chapter 10.1.
I thought my problems was over
but they have yet begun.

But I'm ready for the challenge
even though the obstacles might seem unstoppable.
But with a little hard work and faith
anything is possible.

Thanks to Ms. Robinson and my Sophomore tutor (lol) (she knows who she is) I've almost caught up with the class. And I'm looking forward to learning new things

Formulas I learned this week:
Minutes and seconds to degrees
M/60 + S/3600

Ex: 151 50'27"
50/60+27/3600=150.8408 degrees

Radian to Degrees r= pie
180/r

Ex: 28r/15
180/r*28r/15=336 degrees

Charlie.

this week we reviwed old stuff and took exams.
on friday we started chapter 10 section 1.
we can't use a calculater int his chapter.
first you hae to use the formluas:
cos(alpha+/- beta) = cos(alpha) cos(beta) -/+ sin(alpha) cos(beta)
& sin(alpha +/- beta) = sin(alpha) cos(beta) +/- cos(alpha) sin(beta)
alpha and beta come from the trig. chart and add or subtract to get the angle you are looking for.

one thing is where the problem says "suppose that sin or cos =", in these problems you have to figure out the othe cosines or sines yourself.
For Example:
suppose that sin (alpha) = 4/5 and sin (beta) = 5/13 where 9cosalpha cosbeta - sinalpha sinbeta
cosalpha cosbeta - (4/5)(5/13)
then to find out cosalpha and cosbeta 3/5 and cos beta = -12/13, soo..
(3/5)(-12/13) - (4/5)(5/13)
(-36/65) - (20/65)
= -56/65

Nathan's Blog

In this blog, I will explain chapter 10 section 1. This is the first section of the second nine weeks, and I think everyone, including me, got the hang of this on the first day.

Rule number one(worst rule ever): NO CALCULATOR!!!!

Another good thing to know, is the Trig Chart. This will save you.

Here are the formulas you will need to know:

(cos(alpha)+-(beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)

sin(alpha)+-(beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
This may seem complicated, but it really isn't.

Ex. Find the exact value of sin15

To solve this, you must find two trig chart angles that add or subtract to give you 15.
alpha=45
beta=30
45-30=15

Now, you just plug it into the sin formula.

sin(45-30)=sin45cos30-cos45sin30
=(√2/2)(√3/2)-(√2/2)(1/2)
=√6/2-√2/4=√6-√2/4

sin15=√6-√2/4

Ex.2 Find the exact value of: cos(75)

alpha=45
beta=30

cos(45+30)=cos(45)cos(30)-sin(45)sin(30)
=(√2/2)(√3/2)-(√2/2)(1/2)
=√6/2-√2/4

cos75=√6-√2/4
It's the same answer as the first one if you didn't catch on.

See how easy this section is. Like all of them, this chapter is only going to get harder.

Taylor's 7th(or 8th) Blog

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= 6
B= 65 degrees b=?
C= 25 degrees c=3

First, draw your triangle with the information that I gave you includes.

Cos 25= ?/6 Then, write out your equation like the formula I gave you.

b= 5.438 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= 10
B= ? degrees b=6
C= ? degrees c=7

First, draw your triangle with the information that I gave you includes.

Cos C= 6/10 Then, write out your equation like the formula I gave you.

C= Cos^-1 (6/10) 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 sue 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 -53. 130 degrees= 36.87 degrees Finally, you just have to subtract 90 degrees and 53. 130 degrees from 180 degrees and you will get 36.87 degrees for angle B.

So your answers will be C= 53. 130 degrees and B= 36.87 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)

Nicala's Blog

In chapter nine section two





we learned how to find the area of a triangle.





one divided by two(base)(height) is only for right triangle.





one divided by two(A)(B)(sine angle between) for non-right triangle





example problem





Two sides of a triangle have lengths of eight centimeters and six centimeters. The angle between the two sides measures to 85 degrees. Find the area of the triangle.





First you have to draw your triangle and your triangle is not right so you can have to use the second formula.





A= one/two(A)(B)(sine(theta)



A= one/two(eight)(six)(sine(eighty five)



A= one/two(forty eight)(sine(eighty five)



A= twenty four (sine(eighty five)



A= twenty three. nine centimeters squared

Week 7 Blog Prompt

What concept do you feel like you mastered this nine weeks? What concept did you struggle with the most? Why? What can you change this nine weeks in your study habits, etc to improve your grade?

Dom's blog

Law of Sines. Chapter 9-3.
Law of sines is used for non right triangles and used only when you have an angle and an opposite leg value.
The formula is : sin(angle1)/opp. leg 1 = sin(angle2)/opp. leg 2

In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T.

sin126/12=sinT/7
T=sin-1(7sin126/12)
T=28.159 degrees
Sin is related to Y and is positive in Q I and II.
-28.159+180=151.841
151.841 + 126 is greater than 180. And that equals 28.159.

CHARLIE'S.

This week we didn't really learn new stuff, we just took a test tuesday through friday and reviewed the chapter on monday.
But anyways...
One of the things on the test was law of sines.
Law of sines only works when you have an angle with a side opposite to that angle, and you also need either another angle or another side. So, what you do is set up for the formula...
sintheta/opposite leg = sintheta/opposite leg
... then you cross mulitply to solve for either the missing angle or the missing side
For example:
If you had a triangle ABC and angle A is 28 degrees and side a is 8 and side c is 12.
You plug it into the formula as..
sin 28/8 = sinC/12
Then, by cross multipling you get.
12sin28 = 8sinC
Dividing by 8
you then get 12sin28/8 = sinC
Then you take the inverse
C = sin ^-1 (12 sin28/8)
so angle C = 44.77 degrees

Hoppin' on the bandwagon (Dylan's blog)

9-3

Law of Sines

Law of sines is used for non right triangles and used only when you have an angle and an opposite leg value that you know.

The formula is : sin(angle1)/opp. leg 1 = sin(angle2)/opp. 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 measurements the engineer knows: AB=25m, angleA=110 degrees, angleB=20 degrees. Find AB and BC.

B=20 degrees
|\
|-\
|--\
|---\
|----\
|-----\
-------
A ? C =50 degrees
=110 degrees

sin 50/25=sin 20/AC

ACsin50/sin50=25sin20

AC=25sin20/sin50

AC=11.162 m

(time to skip long process for BC)

BCsin50=25sin110

BC=25sin110/sin50

BC=30.667 m

Kaitlyn's Blog #8

In this blog, i will be explaining 9-3 Law of Sines.



You can only use law of sines when you have a non right triangle and if you have a pair, an angle and an opposite leg.



Formula: Sin(angle)/opposite leg=Sin(angle 2)/opposite leg 2



Example: Triangle ABC has A(angle)=32 B(angle)=81.8 and a=42.9 inches. What is the measure of side c?



1) First you need to find your other angle C. You can do this by simply subtracting your other two angles from 180, since a triangles angles add up to 180.



2) Then you use law of sines to find your side c.



3) sin32/42.9=sin66.2/c



4) cross multiply to get c x sin32=42.9sin66.2



5) then divide sin32 on both sides to get c=74.07



So that is how you find you other side using law of sines.

I barely remember anything from this week so I am just going to go over section 9-3, Law of Sines.


This is only with non-right triangles, and you only use it 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. In triangle ABC, AB= 25; A=110; and B=20. Find AC and BC.
Draw the triangle. You will find the missing angle by 180-110-20=50.
So, sin50/25=sin110/BC.
Cross multiply to get:
BC=25sin110/sin50BC=30.667 m
Now to find AC: sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m

Ex. 2
In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T.
sin126/12=sinT/7T=sin^-1(7sin126/12)T=28.159 degrees

Now you must find another angle. Sin is related to Y and is positive in Q I and II.-28.159+180=151.841

Now you test the angles in the triangle. 151.841 + 126 exceeds 180. So the only angle that can work, is 28.159.


kthanksbye.

In this blog I will be explaining section 9-3 which explains the law of sines.

-First it is only used with non-right triangles.

- only use when you have an angle and opp leg value you know.

Formulas:

- sin(angle)/opp leg = sin (angle2)/opp leg 2

- then you just cross multiply

Ex.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=25 m; angle A=110 degrees and angle B=20 degrees. Find AC and BC.

First you have to draw the triangle,then you will find the missing angle by subtracting 180-110-20=50.

sin50/25=sin110/BC then you cross multiply to get:BC=25sin110/sin50
BC=30.667 m

Now to find AC: sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m

Ex 2. In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T.

sin126/12=sinT/7
T=sin-1(7sin126/12)
T=28.159 degrees

Now you must find another angle. Sin is related to Y and is positive in Q I and II.
-28.159+180=151.841

Then you test the angles in the triangle. 151.841 + 126 exceeds 180. So the only angle that can work, is 28.159.

Nathan's Blog

For this blog, I will show you the stuff from section 9-3, which involves the Law of Sines.

This is only with non-right triangles, and you only use it 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=25 m; angle A=110 degrees and angle B=20 degrees. Find AC and BC.

Draw the triangle. You will find the missing angle by 180-110-20=50.

So, sin50/25=sin110/BC. Cross multiply to get:BC=25sin110/sin50
BC=30.667 m

Now to find AC: sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m

Ex. 2 In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T.

sin126/12=sinT/7
T=sin^-1(7sin126/12)
T=28.159 degrees

Now you must find another angle. Sin is related to Y and is positive in Q I and II.
-28.159+180=151.841

Now you test the angles in the triangle. 151.841 + 126 exceeds 180. So the only angle that can work, is 28.159.

That's all for this blog. Just a little note here. The THETA symbol, is the lowercase y in the WEBDINGS font. Your Welcome.

Helen's Blog

I'm going to show you how to do 9-2.\

Formulas that you can use:

• 1/2 bh only use this for a right triangle

• 1/2 bh sin(angle b/w) only use this formula for a non right triangle

Examples:

Two sides of a triangle have lengths 9cm & 3cm. The angle between the sides measure 64 degrees. find the area of the triangle.

Since there is no right angle you would use: (1/2 bh sin(angle b/w))

Then you would just plug the numbers in.
a=1/2 (9)(3) sin 64 degrees
a=13.5 sin 64 degrees
a=12.1337 cm squared

And that's all you have to do and if you have a triangle with a right angle you would just have to use 1/2 bh. ( This was the easiest thing I learned all week its not that hard as long as you know the formula).

Malorie's Blog

9-4 Law of Cosines

in 9-4, we are learned how to solve triangles using law of cosines.
you use this when you do not have a right angle or you can't use any of the other formulas. Law of Cosines is more like a last resort.

The formula is
(opposite leg) squared= (adjacent leg) squared + (adjacent leg) squared - 2(adjacent leg)(adjacent leg)Cos(angle between)

Example:

Two legs of a triangle are 2 and 6. The angle between the is 100 degrees. Find the other side.

you are looking for the opposite of 100 degrees, you can't use law of sines because you do not have any opposite lengths.

x^2=2^2 + 6^2 - 2(2)(6)Cos 100

To solve this you put the WHOLE equation under the square root thing or your answer wont come out right.

x=6.65


Example 2:

The lengths of a triangle are 3, 4, and 5. Find all angles.

First, we'll find angle A.
(you can find whatever you choose first)

4^2= 5^2 + 3^2 - 2(5)(3)Cos A

4^2 - 5^2 - 3^2= -2(5)(3)Cos A

divide by cos A, then take the inverse of cos to get and angle measure.

A=53.1

now, you can either do the same for the rest of the measures or use law of sines.
I prefer to use law of sines because i now have an opposite.

Sin B/5 = Sin 53.1/4

Cross multiply and then divide 4, then take the inverse of Sin B

B = 88.4

Now that you have two angles you can just subtract them from 180 to get your third angle

180-88.4-53.1 = 38.5

C= 38.5

Lawrences blog again

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 aslongas you know the formulas and you just plug the numbers into them. that is pretty much this whole chapter. you have to know the formulas if you want to pass the chapter. there aren't that many formulas to remember so don't be lazy and not know them. if you do that then you should just quit.

Taylor's 6th Blog (i think lol)

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= 5
B= 20 degrees b=6
C= 70 degrees c=?

First, draw your triangle with the information that I gave you includes.

Sin 70= ?/5 Then, write out your equation like the formula I gave you.

c= 4.698 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= 5
B= ? degrees b=4
C= ? degrees c=3

First, draw your triangle with the information that I gave you includes.

Sin C= 3/5 Then, write out your equation like the formula I gave you.

C= Sin^-1 (3/5) Then, you need to find the inverse.

C= 36. 870 degrees Then, solve and round to the third place after the decimal. 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 143.13 degrees. Then, add 90 degrees to it and you get 233.13. Since this is over 180 degrees we cannot use it for triangles only have 180 degrees in them.

180 degrees -90 degrees -36. 870 degrees= 53.13 degrees Finally, you just have to subtract 90 degrees and 36. 870 degrees from 180 degrees and you will get 53.13 degrees for angle B.

So your answers will be C= 36. 870 degrees and B= 53.13 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)

Nicala's Blog

in chapter nine section 4





we are learning about the law of cosine





the law is used only when you do not have a right triangle or when you can not use law of sine (which i explained in my last blog)





the formula for law of cosine is (opposite leg) squared=(adjacent leg)squared + (adjacent leg)squared-2(adjacent leg)(adjacent leg)cosine(angle between)

example problem

suppose that two side of a triangle have lengths 4 and 8 cm and that angles between them that measures to 120 degrees. find the third side.

first you must draw your triangle so you can have picture of what you are doing.

then you need to solve your problem.

P squared=(4)2+(8)2-2(4)(8)cosine(120)

when you putting your problem in your calculator you need to put whole problem under a square root or you will not get the right answer.

P=10.58

Week 6 Prompt

How is SOHCAHTOA in Ch. 9 and The unit circle in Ch. 7 connected? Give examples of the connection.

dylans blog of stuffs :D

Sec. 9-4
Law of Cosines (opp leg) squared =(adj leg)squared +(adj leg) squared-2(adj leg)(adj leg) cos(angle b/w)

This is your last resort if nothing else works. The formula is easy, but you have to remember all of it.

Ex. triangle w/ 3 cm side and 7 cm side, angle between them is 130 degrees

(z)^2= (3)^2+(7)^2-2(3)(7) cos(130deg)
z=sq, root. 3^2 +7^2- 2(3)(7) cos(130 deg)
z=9.22 cm^2

malorie's blog

9-3 Law of Sines

In order to do law of sines there must be a right angle and there must be an angle with an opposite to use the formula.

Formula:

sin(angle)/ opposite leg = sin(angle 2) opposite leg 2

Example:

in the triangle ABC, angle B= 126, b=12, c=7. Determine if angle C exsist. Find all possible measures of angle C.

sin126/12= sin C/7

126 is oppisite to 12
you would use sinC because C is the angle you are looking for, and 7 is opposite of it.

12 sin C = 7 sin 126 (Cross Multply)

divide by 12 to get sin C by itself.

take the inverse of sin C.

C=28.159

you have to find all values of C so you would find the second quadrant angle becaue sin is positive in the first and second quadrant.

you get 151.841 which isn't reasonable so you don't use it.

your final answer is C= 28.159



Example 2:

Solve the triangle ABC. Angle C= 25, c=2, and b=3.

sin25/2= sinB/3
cross multiply
divide by 2
take the inverse
B=39.3


A= 115.7

sin25/2= sin 115.7/a
cross multiply
divide by sin25
solve



It's that simple (:
a=4.26

Doms blog

Section 9-4. which is the law of cosines.
Law of Cosines formula: (opp leg) squared =(adj leg)squared +(adj leg) squared-2(adj leg)(adj leg) cos(angle b/w)
and law of cosines is only used with isosceles triangles.

Ex: Supposed that two sides of a triangle have lengths 3cm and 7cm and that the angle b/w them measures 130 degrees. find the third side.

(p)squared= (3)squared+(7)squared-2(3)(7) cos(130degrees)
p= square root of 3squared +7squared- 2(3)(7) cos(130) degrees
p=9.22 cmsquared

BLAOW!!!! David's Blog.

Section 9-3 was about Law of Sines, which was kind of easy.

There are two rules that you NEED to know though
First Law of Sines can only be used on non right triangles
Second it can only be used when you have an angle and opposite leg that are known.

Other than that all you need to know is the very simple formula for it.

sin(angle)/opposite leg = sin(angle 2)/opposite leg 2

Just cross multiply and solve.

For instance: A triangle FTL has three angles, F=110 T=20 L=50, and leg FT = 25 meters. Find legs FL and LT.

L50



? ?



F110 20T
25 m


Sorry about the bad visual but I think you get the point.

Okay so first we're going to find FL. Our triangle is not right and we have an angle L and an opposite leg of L so we are good to go.

soo sin(50)/25 sin(20)/FL
cross multiply and we get FL sin 50= 25 sin 20
solve FL = 25 sin 20/sin 50
plug into a calculator FL = 11.162

now for LT

sin(50)/25=sin(110)/LT
LT sin 50= 25 sin 110
LT = 25 sin 110/sin 50
LT = 30.667


Easy enough right?

Mary's Blog :/

to the correct units.

For this 150 words, I will be reviewing section 9-4, Law of Cosines. You use this when you don't have a pair for Law of Sines. B-Rob told us to only use this formula for a LAST resort because its complicated and leaves more room for mistakes.

Here's the formula:

(opp leg)^2=(adj leg)^2+(adj leg)^2-2(adj leg)(adj leg)cos(angle b/w)

you can plug in for any part of this formula and solve.

Ex. you have a triangle with side lengths of 3 and 7 with an angle between that measures 130 degrees. Find the third side.

the first thing you always do is draw out your triangle, then you can see that you cannot use sohcahtoa because it has no information that would lead to it being a right triangle, so you check for law of sines, there isn't a pair for you to use, so you turn to your last resort, law of cosines

you plug in your fomula

you get x^2=3^2(7^2)-2(3)(7)cos(130)

type that in your handy dandy calculator and you got it!

X=9.22 and if they give you units, you would indicate your answer

Kaitlyn's Blog #7

In this blog, i will be going over section 9-4.


This section was easy for me. All you have to do is plug your information into the formula and solve. In this section we learned about the Law of Cosines. This is used when you dont have a pair in a non right triangle.

Formula: opp leg(squared)=adj leg(squared) + adj leg(squared) - 2 (adj leg)(adj leg)

cos(angle b/w)

Example: a non right triangle has sides 5cm and 3cm. The angle between those sides is 50.

First you need to sketch out the triangle to see what you have to work with. You want to find the other side of the triangle. To do this you must plug your information into the formula.

x(squared)=5(squared) + 3(squared) - 2(5)(3) cos(50)
Square root everything to get:
x=[square root of] 5(squared) + 3(squared) - 2(5)(3) cos (50)

Plug this into your calculator exactly as shown and your answer comes out to be: x=3.836 cm^2

This is what i learned form Section 9-4...

Lawremce"s Blog

In this blog i will go over section 9-4. this section is the law of cosines. this section isnt very hard at all. you just have to plug into the formula and make sure you didnt make a mistake on your calculator or you wont get an anser.

* Law of Cosines is used when you dont have a pair in a non right triangle.
formula: (opp leg) squared =(adj leg)squared +(adj leg) squared-2(adj leg)(adj leg) cos(angle b/w)

examples:

1. Supposed that two sides of a triangle have lengths 3cm and 7cm and that the angle b/w them measures 130 degrees. find the third side.

draw a triangle and label it according to what they told you.
then you just plug the number into the formula.

(p)squared= (3)squared+(7)squared-2(3)(7) cos(130degrees)

then you square root everything

p= square root of 3squared +7squared- 2(3)(7) cos(130) degrees

your answer should be:

p=9.22 cmsquared

That is what i learned from section 9-4. its not hard but you just have to learn the formula and you will be fine.

In this blog, I will go over section 9-2, which involves the area of right and non-right triangles.
Formulas

-1/2 bh - only right triangles

-1/2 ab sin(angle between) - only for non-right triangles

Ex. Two sides of a triangle have lengths 7cm and 4cm. The angle between the siddes measures73 degrees. Find the area of the triangle.

A=1/2 ab sin(theta)
A=1/2(7)(4) sin73
A=14sin73
A=13.388 cm squared

Ex2 The area of triangle PQR is 15. p=5 and q=10. Find all possible measures of angle r.

first you have to draw a triangle. And then you have to label the sides bottom left point=P; top point=Q; bottom right point=R

15=1/2(5)(10) sin (theta)
15=25 sin (theta)
sin(theta)=15/25
(theta)=sin^-1(15/25)
(theta)=36.870 degrees

36.870 is in the first quadrant. Sin is positive in Q I and II.
-36.870+180=143.130 degrees.

So theta=36.870 degrees and 143.130 degrees.

Toraayyyy's blogg.

In this blog, I will go over section 9-2, which involves the area of right and non-right triangles.1/2 bh - only right triangles1/2 ab sin(angle between) - only for non-right trianglesEx. This is just a regular triangle. The left side is 7 and the right side is 4. The angle between them is 73 degrees. This is a non-right triangle.A=1/2 ab sin(theta)A=1/2(7)(4) sin73A=14sin73A=13.388 cm squaredEx. 2 The area of triangle PQR is 15. p=5 and q=10. Find all possible measures of angle R.Draw a regular triangle. (An arbitrary one as B-Rob says.)Bottom left point=P; Top point=Q; Bottom right point=R15=1/2(5)(10) sin (theta)15=25 sin (theta)sin(theta)=15/25(theta)=sin^-1(15/25)(theta)=36.870 degrees36.870 is in the first quadrant. Sin is positive in Q I and II.-36.870+180=143.130 degrees.So theta=36.870 degrees and 143.130 degrees.