Trigonometry & Inverse Trigonometry
Heights And Distances
nta_abhyas_2025
Grade None
Question:
When the elevation of the sun changes from $45°$ to $30°$, the shadow of a tower increases by 60 units, then the height of the tower is
30√3 units
30 (√3 + 1) units
30 (√3 - 1) units
30 (√2 + 1) units
Step-by-Step Solution
Key Concept: Combine multiple angle of elevation/depression equations with a known horizontal distance to solve for the height
From the diagram, the observer stands at height $h$ with angles of $45°$ and $30°$ to two points on the ground separated by distance $60$ units. Using $\tan 45° = \frac{h}{a} \Rightarrow h = a$, and $\tan 30° = \frac{h}{a+60} \Rightarrow \frac{h}{\sqrt{3}} = a + 60$. Substituting $h = a$: $\frac{a}{\sqrt{3}} = a + 60 \Rightarrow a(1 - \sqrt{3}) = 60\sqrt{3} \Rightarrow a = 30(\sqrt{3} + 1)$. Therefore $h = 30(\sqrt{3} + 1)$ units.
Correct Answer: 30(√3 + 1)