Medium
The angles of elevation of the top of a tower from two points at a distance of $4 \,m$ and $9 \,m$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6 \,m$.
(N/A) Let $AQ$ be the tower and $R, S$ be the points at a distance of $4 \,m$ and $9 \,m$ from the base of the tower $(Q)$ respectively.
The angles of elevation are complementary. Therefore,if one angle is $\theta$,the other will be $(90^\circ - \theta)$.
In $\triangle AQR$,
$\tan \theta = \frac{AQ}{QR} = \frac{AQ}{4} \quad \dots(i)$
In $\triangle AQS$,
$\tan(90^\circ - \theta) = \frac{AQ}{SQ}$
$\cot \theta = \frac{AQ}{9} \quad \dots(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \left(\frac{AQ}{4}\right) \cdot \left(\frac{AQ}{9}\right)$
$1 = \frac{AQ^2}{36}$
$AQ^2 = 36$
$AQ = \sqrt{36} = 6 \,m$ (Since height cannot be negative).
Hence,the height of the tower is $6 \,m$.
The angles of elevation are complementary. Therefore,if one angle is $\theta$,the other will be $(90^\circ - \theta)$.
In $\triangle AQR$,
$\tan \theta = \frac{AQ}{QR} = \frac{AQ}{4} \quad \dots(i)$
In $\triangle AQS$,
$\tan(90^\circ - \theta) = \frac{AQ}{SQ}$
$\cot \theta = \frac{AQ}{9} \quad \dots(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \left(\frac{AQ}{4}\right) \cdot \left(\frac{AQ}{9}\right)$
$1 = \frac{AQ^2}{36}$
$AQ^2 = 36$
$AQ = \sqrt{36} = 6 \,m$ (Since height cannot be negative).
Hence,the height of the tower is $6 \,m$.
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