Question

Question 14
From the top of a tower h m high angles of depression of two objects, which are in line with foot of the tower are $\alpha$ and $\beta (\beta >\alpha )$. Find the distance between the two objects.

Answer 1

Let the distance two objects is x m.

And CD = y m

Given that, $\angle BAX=\alpha =\angle ABD$, [alternate angle]

$\angle CAY=\beta =\angle ACD$ [alternate angle]

And the height of tower, AD = hm

Now, in $\Delta ACD$,

$tan\beta =\frac{AD}{CD}=\frac{h}{y}$

$\Rightarrow y=\frac{h}{tan\beta }\dots \dots \left(i\right)$

And in $\Delta ABD$,

$tan\alpha =\frac{AD}{BD}\Rightarrow \frac{AD}{BC+CD}$

$\Rightarrow tan\alpha =\frac{h}{x+y}\Rightarrow x+y=\frac{h}{tan\alpha }$

$\Rightarrow y=\frac{h}{tan\alpha }-x$

From Eqs (i) and (ii),

$\frac{h}{tan\beta }=\frac{h}{tan\alpha }-x$

$\therefore x=\frac{h}{tan\beta }=\frac{h}{tan\alpha }-x$

$=h(\frac{1}{tan\alpha }-\frac{1}{tan\beta })=h(cot\alpha -cot\beta )[\because cot\theta =\frac{1}{tan\theta }]$

Which is the required distance between the two objects.