Solving Trigonometry Problems: How to Find CE Using the Cosine Rule

[terryds]

Homework Statement

I don't understand how to get CE.

Homework Equations

Trigonometric identities (maybe)

The Attempt at a Solution

[/B]
First, I start with the cosine rule
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 going to be a quadratic equation and it's hard to obtain what CE is (using this method)
Please help

 

[Simon Bridge]

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

 

[PPHT123]

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

 

[terryds]

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

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

 

[terryds]

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

CB/ sin alpha = CA / sin beta

CE = CA sin alpha
CE = CB sin beta

Then whatt??

 

[PPHT123]

CB/ sin alpha = CA / sin beta

CE = CA sin alpha
CE = CB sin beta

Then whatt??

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

 

[terryds]

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

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 ?

 

[PPHT123]

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 ?

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

 

[terryds]

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

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

 

[Ray Vickson]

Homework Statement

I don't understand how to get CE.

Homework Equations

Trigonometric identities (maybe)

The Attempt at a Solution

[/B]
First, I start with the cosine rule
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 going to be a quadratic equation and it's hard to obtain what CE is (using this method)
Please help

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