Question

The angle of elevation of the top of a tower standing on a horizontal plane from two on a line passing through the foot of the tower at a distance 9 ft and 16 ft respectively are complementary angles. Then the height of the tower is:
A 9 ft
B 12 ft
C 16 ft
D 144 ft

Answer 1

Hint: In this problem, first we need to draw the given situation. Next, apply the trigonometric identities to obtain the relation between the elevation angle and height of the triangle.

Complete step-by-step answer:
Consider $AB$ to be the height of the tower. The angle of elevation at a distance of 9ft is $\theta$ and angle of elevation at a distance of 16 ft be $90 - \theta$, because both the angles are complementary as shown below.

Now, in $\Delta ABC$,

\,\,\,\,\,\,\cot \theta = \dfrac{{BC}}{{AB}} \\\ \Rightarrow \cot \theta = \dfrac{9}{{AB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\\ \end{gathered}$$ In $$\Delta ADB$$, $$\begin{gathered} \,\,\,\,\,\,\tan \left( {90 - \theta } \right) = \dfrac{{AB}}{{BD}} \\\ \Rightarrow \cot \theta = \dfrac{{AB}}{{16}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\\ \end{gathered}$$ From, equation (1) and (2). $$\begin{gathered} \,\,\,\,\,\,\dfrac{9}{{AB}} = \dfrac{{AB}}{{16}} \\\ \Rightarrow A{B^2} = 9 \times 16 \\\ \Rightarrow A{B^2} = 144 \\\ \Rightarrow AB = \sqrt {144} \\\ \Rightarrow AB = 12ft \\\ \end{gathered}$$ Thus, the height of the tower is 12 ft, hence, option (B) is the correct answer. Note: In this problem, both the elevation angles are complementary to each other, hence, the sum of the elevation angles will be the right angle.