# Trigonometry problem

1. Oct 7, 2016

### terryds

1. The problem statement, all variables and given/known data

I don't understand how to get CE.

2. Relevant equations
Trigonometric identities (maybe)

3. The attempt at a solution

CB^2 = AC^2 + AB^2 - 2 AC AB cos α
(CE/sin(β))^2 = (CE/sin(α))^2 + L^2 - 2 (CE/sin(α)) L cos α

which is gonna be a quadratic equation and it's hard to obtain what CE is (using this method)

2. Oct 7, 2016

### Simon Bridge

Try the usual way you derive the cosine rule - except solving for the altitude of the triangle instead of factoring it out.

3. Oct 7, 2016

### PPHT123

I think using sine rule to find a side first is a faster way

4. Oct 7, 2016

### terryds

I get this $(\frac{1}{(sin^2(\beta))})CE^2+(\frac{2\ L\cos \alpha}{sin \ \alpha})CE-L^2=0$

Then, use the quadratic formula to solve CE??
But if I use quadratic formula, there will be two solutions, but there is only one CE.
And I doubt if the solution obtained will be the same as in the photo I gave

5. Oct 7, 2016

### terryds

CB/ sin alpha = CA / sin beta

CE = CA sin alpha
CE = CB sin beta

Then whatt??

6. Oct 7, 2016

### PPHT123

angle ACB = 180 - a -b
you can try to use L/sin(180-a-b) = ??

7. Oct 7, 2016

### terryds

Angle ACB = 180 - (a + b)
sin(180-(a+b)) = sin (a+b)

L/sin(a+b) = CB/ sin (a) = CA / sin (b)

Then??
How to make CE (the altitude) show up ?

8. Oct 7, 2016

### PPHT123

You succeeded to express CB (or CA) in terms of L , a and b. Then what is CB sin(b) ?

9. Oct 7, 2016

### terryds

L/sin(a+b) = CE/sin(a)Sin(b)
CE = L sin(a) sin(b)/ sin(a+b)

Thanks a lot for your help bro

10. Oct 7, 2016

### Ray Vickson

$AE/CE = \cot(\alpha)$ and $BE/CE = \cot(\beta)$. Now add and manipulate.