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Introduction to Trigonometry and Its Applications
NCERT Exemplar
CBSE
Grade 10
Question:
The angle of elevation of the top of a building from the foot of a tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is $50\text{ m}$ high, find the height of the building.
Step-by-Step Solution
Key Concept: Tower height $= 50\text{ m}$. In tower triangle: $\tan 60^\circ = \dfrac{50}{d} \Rightarrow d = \dfrac{50}{\sqrt{3}}$. In building triangle: $\tan 30^\circ = \dfrac{h}{d}$.
Distance between foot of building and tower $= d$. In right triangle of tower: $\tan 60^\circ = \dfrac{50}{d} \Rightarrow \sqrt{3} = \dfrac{50}{d} \Rightarrow d = \dfrac{50}{\sqrt{3}}\text{ m}$. [1.0 Mark] In right triangle of building: $\tan 30^\circ = \dfrac{h}{d} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{50/\sqrt{3}} \Rightarrow h = \dfrac{50}{3} = 16.67\text{ m}$ (or $16\dfrac{2}{3}\text{ m}$). [1.5 Marks] Height of building is $16\dfrac{2}{3}\text{ m}$. [0.5 Mark]
--- 🎯 Official CBSE Marking Scheme: Finding distance $d = 50/\sqrt{3}\text{ m}$: 1.0 Mark Setting up building angle equation $\tan 30^\circ = h/d$: 1.5 Marks Calculating height $h = 16\frac{2}{3}\text{ m}$: 0.5 Mark
Correct Answer:
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