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

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So:

(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)

tan (alpha) = h/CO

CO = h/tan (alpha)

CO = h*cot (alpha)

working out that in triangle CBO the angle is unkown called theta and triangle ABO the angle must be 180 - theta

(iii) using cosine law with Cos (theta as the subject)

CBO:

Cos (theta) = {(2x)^2 + (BO)^2 - (CO)^2}/ 2*(2x)*(BO)

ABO:

Cos (180 -(theta)) = {(x)^2 + (BO)^2 - (AO)^2}/ 2*(x)*(BO)

Therefore, Cos (180 -(theta)) = -cos (theta)

therefore {(O) - (BO)^2 - (x)^2}/ 2*(x)*(BO)

Replacing AO,BO & CO for the cot values these two equations shouldbe equal to one another. I don't seem to get the right answer after manipulating them though

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