Thread: Trig equation

Hi
Can anyone tell me how to do this:

If cosecant A=3.168 find secant A

-I understand that secant=1/cos and that cosecant is 1/sin but I am still confused
Thanks a lot to anyone who took the time to read this!

Originally Posted by stephie

Hi
Can anyone tell me how to do this:

If cosecant A=3.168 find secant A

-I understand that secant=1/cos and that cosecant is 1/sin but I am still confused
Thanks a lot to anyone who took the time to read this!

ok, so you know the relationships, so use them. here's how

i will use csc for cosecant and sec for secant

cscA = 3.168
=> 1/sinA = 3.168
=> sinA = 1/3.168 = 0.316
now we know that sin^2(A) + cos^2(A) = 1
we know the value of sinA so let's plug it in to find cosA

=> (0.316)^2 + cos^2(A) = 1
=> cos^2A = 1 - (0.316)^2 = 0.9
=> cosA = 0.949

so cosA = 0.949
=> 1/cosA = 1/0.949 = 1.054
=> secA = 1.054....

i rounded off some stuff while doing this problem, if you want a more accurate answer you can redo this with more decimal places

thank you

Hello, stephie!

If csc A = 3.168, find sec A.

. . . . . . . . . . . . . . . - . . . . . . . . . . __
That decimal looks suspiciously like √10.
. . . . . . . . . . . . . . . . __
. . . . . . . . . . . . . . . √10 . - . hyp
We have: . csc A .= .------ .= .-----
. . . . . . . . . . . . . . . . 1 . - . . opp

. . . . . . . . . . . . . . . . . . . . . . . . - . . . - . . . . . . . . __
Hence, A is an angle in a right triangle with: .hyp = √10 and opp = 1.


Using Pythagorus, we find that: .adj = 3

. . . . . . . . . . . . . . . . . . . . . . .__
. . . . . . . . . . . . . . . .hyp . . . √10
Therefore: . sec A .= .----- .= .-----
. . . . . . . . . . . . . - . .adj . . . . .3

Originally Posted by Soroban

Hello, stephie!

. . . . . . . . . . . . . . . - . . . . . . . . . . __
That decimal looks suspiciously like √10.
. . . . . . . . . . . . . . . . __
. . . . . . . . . . . . . . . √10 . - . hyp
We have: . csc A .= .------ .= .-----
. . . . . . . . . . . . . . . . 1 . - . . opp

. . . . . . . . . . . . . . . . . . . . . . . . - . . . - . . . . . . . . __
Hence, A is an angle in a right triangle with: .hyp = √10 and opp = 1.


Using Pythagorus, we find that: .adj = 3

. . . . . . . . . . . . . . . . . . . . . . .__
. . . . . . . . . . . . . . . .hyp . . . √10
Therefore: . sec A .= .----- .= .-----
. . . . . . . . . . . . . - . .adj . . . . .3

well, 3.168 is a bit bigger than sqrt(10), but i guess that will suffice. your way is a much nicer way to do the problem i would have done it that way if they gave us a whole number or fraction