Question 11 Mark
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^o.$
Answer
First, let us draw a simple diagram to represent the problem. Here $A B$ represents the tower, $C B$ is the distance of the point from foot of the tower and $\angle A C B$ is the angle of elevation. We need to determine the height of the tower, i.e., $A B$. Also, $A C B$ 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 $A B$ and $B C$.
Now, $\tan 60^{\circ}=\frac{ AB }{ BC }$
i.e., $\sqrt{3}=\frac{ AB }{15}$
i.e., $A B=15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3} m$
View full question & answer→First, let us draw a simple diagram to represent the problem. Here $A B$ represents the tower, $C B$ is the distance of the point from foot of the tower and $\angle A C B$ is the angle of elevation. We need to determine the height of the tower, i.e., $A B$. Also, $A C B$ 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 $A B$ and $B C$.
Now, $\tan 60^{\circ}=\frac{ AB }{ BC }$
i.e., $\sqrt{3}=\frac{ AB }{15}$
i.e., $A B=15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3} m$