Trigonometry & Inverse Trigonometry
Heights And Distances
nta_abhyas_2025
Grade 11
Question:
A vertical tower subtends an angle of $60°$ at a point on the same level as the foot of the tower. On moving $100$ m further from the first point in line with the tower, it subtends an angle of $30°$ at the point. If the height of the tower is $H$ m, then the value of $\frac{H}{\sqrt{3}}$ (in meters) is
Step-by-Step Solution
Key Concept: In a triangle with two known angles and a known base, use tangent ratios from each angle to find the height and position.
Given a right triangle with angles $60°$ and $30°$, and the base distance is $100$ m. Using $\tan 60° = \frac{H}{x}$ and $\tan 30° = \frac{H}{100-x}$, we have $H = x\sqrt{3}$ and $H = \frac{100-x}{\sqrt{3}}$. Equating: $x\sqrt{3} = \frac{100-x}{\sqrt{3}}$, so $3x = 100 - x$, giving $x = 25$. Therefore $H = 25\sqrt{3}$ m. The ratio $\frac{x}{H} = \frac{25}{25\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$, which simplifies to $9$.
Correct Answer: 2