Question

Trigonometry help, please? I've tried solving this for the third time and haven't gotten it right :/ I keep getting 16/63. Picture of question in discussion post below.

Answer 1

make sure you in degrees mode

Answer 2

I'm sorry, what do you mean @timo86m ?

Answer 3

calcs have 2 modes degree and radian make sure you in degree

Answer 4

Oh, how do I do that? (sorry if it's obvious)

Answer 5

@jim_thompson5910 I'm sorry to bother you, but you helped me before with a question like this and I was hoping you could help me again? I did what you showed me last time, but my teacher says my answer is wrong.

Answer 6

16/63 i get

Answer 7

Me too! hmm... maybe my teacher us wrong?

Answer 8

is the computer or teacher saying otherwise?

Answer 9

Well, the online test marked my answer (16/63) as incorrect and my teacher said to just try and solve for it again :p

Answer 10

maybe cuzz of the specifications it gives you it wanst a somewhere in 2nd quad and b in 3rd quad

Answer 11

the calcs arcsin and arc cos all give you an answer in different quad maybe

Answer 12

I'm not that great at this so I'm not sure :/ I tried solving for the missing length of SinA triangle and than kept tan B (-4/3) since it was tangent and solved with those

Answer 13

Did you use the identity

tan(A+B) = (tan(A) + tan(B))/(1 - tan(A)*tan(B)) ?

Answer 14

Yes, tan A being 12/5 and tan B being -4/3

Answer 15

remember, A is between 90 and 180, so tan(A) is negative

Answer 16

(12/5) + (-4/3) all divided by 1 - (12/5)(-4/3)

Answer 17

since cos is negative here

Answer 18

Ah, right. That part always throws me :p

Answer 19

and sine is positive

Answer 20

So would it look something like this? (-12/5) + (-4/3) all divided by 1 - (-12/5)(-4/3)

Answer 21

you got it

Answer 22

Awesome I'll check it right now (: Thank you again!

Answer 23

np

Answer 24

let me know how it goes

Answer 25

Alrighty (:

Answer 26

56/33 @jim_thompson5910 Thank you (: Lol And thank you too @timo86m

Answer 27

you're welcome, glad it worked