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

  • Thread starter Thread starter terryds
  • Start date Start date
  • Tags Tags
    Trigonometry
AI Thread Summary
To find CE using the cosine rule, the discussion emphasizes starting with the cosine rule formula CB^2 = AC^2 + AB^2 - 2ACAB cos α. It suggests deriving the cosine rule while focusing on the altitude of the triangle instead of factoring it out, which complicates the quadratic equation. The sine rule is proposed as a more efficient method, allowing for expressions like CE = L sin(a) sin(b) / sin(a+b). Participants highlight the importance of manipulating relationships between angles and sides to isolate CE effectively. The conversation concludes with a focus on using trigonometric identities to simplify the problem.
terryds
Messages
392
Reaction score
13

Homework Statement



71srah.png


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
 
Physics news on Phys.org
Try the usual way you derive the cosine rule - except solving for the altitude of the triangle instead of factoring it out.
 
I think using sine rule to find a side first is a faster way
 
Simon Bridge said:
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
 
PPHT123 said:
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??
 
terryds said:
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) = ??
 
PPHT123 said:
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 o_O?
 
terryds said:
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 o_O?
You succeeded to express CB (or CA) in terms of L , a and b. Then what is CB sin(b) ?
 
PPHT123 said:
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
 
  • Like
Likes Simon Bridge
  • #10
terryds said:

Homework Statement



71srah.png


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.
 
Back
Top