Lawrence's Blog

the notes i will go over are a combination of sections 8-1 and 8-2.

these are the formulas B-Rob gave us

m=tan(alpha) for a line
tan2(alpha)= B/a-c if A does not =C for a conic
(alpha)= pi/4 if A=C for a conic
ax^2 +bxy+ cy^2+ dx+ey+f=0

example.

to the nearest degree, find the inclination of the line 2x+5y=15

you have to solve for y
5y=-2x+15

then you divide by 5
y=-2/5x+3

now you are able to use one of the formulas b-rob gave us.

the formula that worked in this problem is:
m=tan(alpha)

m is -2/5. so you get:
-2/5=tan(alpha)

now you take the inverse of tan
(alpha)= tan^-1(2/5)= 21.801

now your back into section 8-1 where you have to find what quadrants tan is positive in. tan is y/x. so since you have to find where y and x are positive you look on the chart and its in Q2 and Q4.

now you can continue solving the problem.

Q2: make negative and add by 180 degrees
-21.801+18= 159.801 degrees

Q4: make negative and add by 360 degrees
-21.801+360= 338.199 degrees.

that is your two final answers. i hope yall understood what i did and i also hope it helped those who dont understand it as well. it took me awhile to understand it. i just had to keep working problems then i finally got it.