Standing 45 m from a radio mast on flat ground, a surveyor measures the angle to its tip as 60 deg; what is the mast's height?

45*sqrt(3) m. tan 60 = height/45, so height = 45*sqrt(3) m (since tan 60 = sqrt(3)).

Across a field 90 m wide, a fisherman views the lamp of a lighthouse at an elevation of 30 deg; how high is the lamp above his eye level?

30*sqrt(3) m. height = 90 * tan 30 = 90/sqrt(3) = 30*sqrt(3) m.

A delivery rider parked 50 m from a telecom tower tilts his gaze 60 deg upward to its antenna; how far above the ground is the antenna?

50*sqrt(3) m. height = 50 * tan 60 = 50*sqrt(3) m.

Class 10 · Maths

Some Applications of Trigonometry MCQs

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Heights and distances

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Q1. Standing 45 m from a radio mast on flat ground, a surveyor measures the angle to its tip as 60 deg; what is the mast's height?
- A.22.5 m
- B.15*sqrt(3) m
- C.45 m
- D.45*sqrt(3) m
Answer: D. 45*sqrt(3) m
tan 60 = height/45, so height = 45*sqrt(3) m (since tan 60 = sqrt(3)).
Q2. Across a field 90 m wide, a fisherman views the lamp of a lighthouse at an elevation of 30 deg; how high is the lamp above his eye level?
- A.90 m
- B.90*sqrt(3) m
- C.30*sqrt(3) m
- D.45 m
Answer: C. 30*sqrt(3) m
height = 90 * tan 30 = 90/sqrt(3) = 30*sqrt(3) m.
Q3. A delivery rider parked 50 m from a telecom tower tilts his gaze 60 deg upward to its antenna; how far above the ground is the antenna?
- A.50*sqrt(3) m
- B.50 m
- C.25 m
- D.16.67*sqrt(3) m
Answer: A. 50*sqrt(3) m
height = 50 * tan 60 = 50*sqrt(3) m.