Two points $ A $ and $ B $ are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are $ 60^{\circ} $ and $ 45^{\circ} $ respectively. If the height of the tower is $ 15 \mathrm{~m} $, then find the distance between these points.
Given:
Two points \( A \) and \( B \) are on the same side of a tower and in the same straight line with its base.
The angles of depression of these points from the top of the tower are \( 60^{\circ} \) and \( 45^{\circ} \) respectively.
The height of the tower is \( 15 \mathrm{~m} \).
To do:
We have to find the distance between the points.
Solution:
Let $CD$ be the height of the tower and $A, B$ be the points whose angles of depression are \( 60^{\circ} \) and \( 45^{\circ} \) respectively.
From the figure,
$\mathrm{CD}=15 \mathrm{~m}, \angle \mathrm{DAC}=60^{\circ}, \angle \mathrm{DBC}=45^{\circ}$
Let the between the point $A$ and the foot of the tower be $\mathrm{AC}=x \mathrm{~m}$ and the distance between the point $B$ and the foot of the tower be $\mathrm{BC}=y \mathrm{~m}$.
This implies,
$\mathrm{BC}=y-x \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DC }}{BC}$
$\Rightarrow \tan 45^{\circ}=\frac{15}{y}$
$\Rightarrow 1=\frac{15}{y}$
$\Rightarrow y=15 \mathrm{~m}$..........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DC }}{AC}$
$\Rightarrow \tan 60^{\circ}=\frac{15}{x}$
$\Rightarrow \sqrt3=\frac{15}{x}$
$\Rightarrow x=\frac{15}{\sqrt3} \mathrm{~m}$
$\Rightarrow x=\frac{5\sqrt3(\sqrt3)}{\sqrt3} \mathrm{~m}$
$\Rightarrow x=5(1.732) \mathrm{~m}$
$\Rightarrow x=8.66 \mathrm{~m}$........(ii)
From (i) and (ii), we get,
$y-x=15-8.66 \mathrm{~m}$
$\Rightarrow y-x=6.34 \mathrm{~m}$
Therefore, the distance between the points is $6.34 \mathrm{~m}$.
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