Question

The angle of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at a distance $9ft$ and $16ft$ respectively are complementary angles. Then the height of the tower is:

• A. $9ft$
• B. $12ft$
• C. $16ft$
• D. $144ft$

Answer 1

The correct option is B $12ft$

$AB$ is the tower. Angle of elevation at a distance of $9$ft at point $C=\theta$ and that of angle of elevation at a distance of $16$ft at the point $D$ is ${90}^{\circ }-\theta$

In $△ABC,\cot \theta =\frac{9}{AB}$

In $△ADB,\tan ({90}^{\circ }-\theta )=\frac{AB}{16}$

$\Rightarrow \cot \theta =\frac{AB}{16}$

From above we have $\cot \theta =\frac{9}{AB}=\frac{AB}{16}$

$\Rightarrow \frac{9}{AB}=\frac{AB}{16}$

$\Rightarrow {\left(AB\right)}^2=9\times 16$

$\therefore AB=\sqrt{9\times 16}=3\times 4=12$ft