Question 12 Marks
A tower stands vertically on the ground. from a point on the ground which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$, find the height of the tower.
Answer
View full question & answer→First let us draw a simple diagram to represent the problem (see Fig. 9.4). Here AB represents the tower, CB is the distance of the point from the tower and $\angle ACB$ is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angled at B .
To solve the problem, we choose the trigonometric ratio $\tan 60^{\circ}$ (or $\cot 60^{\circ}$ ), as the ratio involves AB and BC .
Now,$\quad$$\quad$$\tan 60^{\circ}=\frac{AB}{BC}$
i.e.,$\quad$$\quad$$\sqrt{3}=\frac{AB}{15}$
i.e.,$\quad$$\quad$$AB=15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3} m$.
To solve the problem, we choose the trigonometric ratio $\tan 60^{\circ}$ (or $\cot 60^{\circ}$ ), as the ratio involves AB and BC .
Now,$\quad$$\quad$$\tan 60^{\circ}=\frac{AB}{BC}$
i.e.,$\quad$$\quad$$\sqrt{3}=\frac{AB}{15}$
i.e.,$\quad$$\quad$$AB=15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3} m$.