Question

A person observes the elevation of the tower to be theta . On advancing 'p' meters towards the tower the elevation is 45 degree and on advancing 'q' meters nearer the angle of elevation is (90degree-theta) . Prove that the height of the tower is pq/p-q

Answer 1

Refer to the attached image.

Here, let us observe that a person observes the elevation of the tower AB (say of height 'h' meters) to be theta. So, $\angle ACB = \theta$.

On advancing 'p' meters towards the tower that is CD = 'p' the elevation is 45 degree that is $\angle ADE = 45^\circ$.

Then, On advancing 'q' meters that is DE='q' nearer the angle of elevation is $(90^\circ-\theta)$. So, $\angle AEB = (90^\circ-\theta)$ and distance BE = x-(p+q) meters.

We have to prove that the height of the tower = $h = \frac{pq}{p-q}$

Proof:

Consider triangle ABC,

So, $\tan \theta = \frac{P}{B} = \frac{AB}{CB}= \frac{h}{x}$ (Equation 1)

In triangle ADB,

$\tan 45^\circ = \frac{AB}{DB} = \frac{h}{q+x-(p+q)} = \frac{h}{x-p}$

So, h = x-p

Consider triangle AEB,

$\tan(90- \theta) = \frac{AB}{EB}=\frac{h}{x-(p+q)}$

$\cot \theta =\frac{h}{x-(p+q)}$ (Equation 2)

Multiplying equation 1 by 2, we get

$\tan \theta \times \cot \theta = \frac{h}{x} \times \frac{h}{x-p-q}$

$\frac{1}{\cot \theta} \times \cot \theta = \frac{h}{x} \times \frac{h}{h-q}$

$1 = \frac{h}{x} \times \frac{h}{h-q}$

$1 = \frac{h^2}{x(h-q)}$

${h^2}={x(h-q)}$

$(h+p)(h-q) = h^2$

$h^2-hq+ph-pq = h^2$

$h(p-q) = pq$

$h = \frac{pq}{p-q}$

therefore, the height of the tower is $h = \frac{pq}{p-q}$.

Hence, proved.