# Applied tr!ig question

• Feb 4th 2010, 07:20 AM
dojo
Applied tr!ig question
Q: The points A,B & C lie in a straight line with AB=x & BC=2x. A vertical tower OH is of height h and its base O lies in the same horizontal plane as ABC but not on the line ABC.
The angle of elevation of H from A,B & C are alpha, beta & alpha respectively.
(i) prove thah h^2(cot^2(alpha) - cot^2(beta)) = 2x^2

I'm stuck and cant find another worked example of this knid anywhere. Can you help??
• Feb 4th 2010, 07:27 AM
dojo
My workings so far...
Quote:

Originally Posted by dojo
Q: The points A,B & C lie in a straight line with AB=x & BC=2x. Avertical tower OH is of height h and its base Olies in the same horizontal plane as ABC but not on the line ABC.
The angle of elevation of H from A,B & C are alpha, beta & alpha respectively.
(i) prove thah h^2(cot^2(alpha) - cot^2(beta)) = 2x^2

I'm stuck and cant find another worked example of this knid anywhere. Can you help??

(i) I start by taking the triangle AOB where AB = x and AHO & BHO
(ii)Because I'm required to prove using cot I'm going to use tan. Therefore:
tan (alpha) = h/AO
AO = h/tan (alpha)
AO = h*cot (alpha)
tan (beta) = h/BO
BO = h/tan (beta)
BO = h*cot (beta)
(iii) uisng cosine law:
x^2 = (AO)^2 + (BO)^2 - 2(AO)(BO)Cos (theta)
I try to manipluate this to equal 2x^2 but can't see the link.
Can you show me how this is proven? There are other parts to the question but I think once I understand this I will be okay on my own.
many thanks!!