Cosine rule problem

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

Reply 1

ghostwalker

dojos

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

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

Assuming I've corrected the odd typo and what you've got is as in the "quote" above, then it does work out, and you've made a slip in the susequent manipulation somewhere.

First step gives me:

{(2x)^2 + (BO)^2 - (CO)^2}/ 2 = (AO)^2 - (BO)^2 - (x)^2

So,

4x^2 + (BO)^2 - (CO)^2 = 2(AO)^2 - 2(BO)^2 - 2(x)^2

and

6x^2 = 2(AO)^2 +(CO)^2 - 3(BO)^2

Then,

6x^2 = 2 h^2*cot^2 (alpha) +h^2*cot^2 (alpha) - 3h^2*cot^2 (beta)

....

Reply 2

dojos

Thanks so much Ghostwalker - I just couldn't see the wood through the trees! I think I'll practise my algebra.