# Thread: Applied cosine rule problem.

1. ## Applied 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
------------------------------------------------------------------
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

2. Hello, dojo!

Sorry, I don't understand the problem.
Some information is missing . . .

The points $\displaystyle A,B,C$ lie in a straight line with: .$\displaystyle AB=x,\;BC=2x$
. .
Is line AC horizontal?

A vertical tower $\displaystyle OH$ is of height $\displaystyle h$ and its base $\displaystyle O$
lies in the same horizontal plane as $\displaystyle ABC$, but not on the line $\displaystyle ABC.$ .
Really?

The angle of elevation of $\displaystyle H$ from $\displaystyle A,B,C$ are $\displaystyle \alpha,\;\beta,\;\alpha$, resp.

(i) Prove that: .$\displaystyle h^2\left(\cot^2\alpha - \cot^2\beta\right)$ .
. . . equals WHAT?

Assuming $\displaystyle AC$ is horizontal, this is all I get:

Code:
                      H
o
**| *
* * |   *
*  *  |h    *
*   *   |       *
*    *    |         *
*     *     *O          *
* α    * β               α  *
o - - - o - - - - - - - - - - - o
A   x   B          2x           C

3. Ah, yes good point!! the proof is to equal 2x^2 - I have added this to the question now. ABC is a prependicular to O, so in the same plane. Your diagram is correct.

Do my workings seem correct?

D