# Thread: Trigonometry Two Unknown Values

1. ## Trigonometry Two Unknown Values

I have a problem with this question, Can any tell me if i did anything wrong in this question? and give me clues on how to proceed from where i left off

tan(32)=y/d
y=d*(tan(32)) Equation 1

tan(69)=(y+7.8)/d
y+7.8=d*tan(69)
y=d*(tan(69)) - 7.8 Equation 2

Sub equation 1 into equation 2

d*tan(32) = d*(tan(69)) - 7.8

That's where i'm stuck at the moment, i don't really know if my working out is correct, can anyone help me?

I'm trying to find "d" by the way

2. I would use * for multiplication, not x; or you can use implied multiplication or parentheses. Your notation is, I think, messing you up.

3. Thank you, i didn't know that before

By the way, could you help me with this problem? I'm stuck, and don't know if my working out is right.

4. You're welcome. I am trying to help you. Your notation change is definitely helpful. From where you left off, I would gather all terms with d's in them to one side and factor it out. Then what do you think should happen next?

5. Would this be the right working out?

d*tan(32) = d*(tan(69)) - 7.8

(d*tan(32)) - (d*tan(69))=-7.8

d(tan(32)-tan(69))=-7.8

d=-7.8/(tan(32)-tan(69))

6. Perfect. Some people would change the signs on your RHS to simplify a bit further. However, that does appear to me to be the correct answer. Good job!

7. Thank You! Just one last question, for this kind of trigonometry question where i have to unknown values, is there like some kind of name for it? Like finding an unknown using substituion or something? Because i think i should do more of these questions.

8. Sure. I would say you had two equations in two unknowns. In this case, your equations were linear, so you could have used either substitution or elimination. You used substitution (that was where you set the two equations equal to y equal to each other).