Question

The angle of elevation of the top of a tower from a point A due south of the tower is $\alpha$ and from B due east of th tower is $\beta$. If AB = d, show that the height of the tower is $\frac{d}{\sqrt{co{t}^2\alpha +co{t}^2\beta }}$ [3 MARKS]

Answer 1

Let OP be the tower and let A and B be two point due south an deast respectively of the tower such that $\angle OAP=\alpha$ and $\angle OBP=\beta$. Let OP = h.



In $\Delta OAP$, we have

$tan\alpha =\frac{h}{OA}$

$\Rightarrow OA=hcot\alpha$ ....... (i)

In $\Delta OBP$, we have

$tan\beta =\frac{h}{OB}$

$\Rightarrow OB=hcot\beta$ ......... (ii)

Since OAB is a right - angled triangle

$\therefore A{B}^2=O{A}^2+O{B}^2$

$\Rightarrow d^2=h^{2}co{t}^2\alpha +h^{2}co{t}^2\beta$

$\Rightarrow h=\frac{d}{\sqrt{co{t}^2\alpha +co{t}^2\beta }}$

[$\frac{1}{2}$ MARK]

$\begin{array}{c}In\Delta PAQ\\ tan\alpha =\frac{h}{X}\\ X=\frac{h}{tan\alpha }\dots \dots \left(1\right)\\ =hcot\alpha \left[1MARK\right]\end{array}|\begin{array}{c}In\Delta PBQ\\ tan\beta =\frac{h}{y}\\ y=\frac{h}{tan\beta }\\ =hcot\beta \left[1MARK\right]\end{array}$

$x^2+y^2=d^2$ (Pythagoras theorem)

$\therefore d=\sqrt{h^2(co{t}^2\alpha +co{t}^2\beta )}$

$d=h\sqrt{co{t}^2\alpha +co{t}^2\beta }$

$h=\frac{d}{\sqrt{co{t}^2\alpha +co{t}^2\beta }}$ [1.5 MARK]