A 1.5 m tall boy stands some distance from a 30 m tall building. As he walks towards the building, the angle of elevation of its top from his eyes changes from 30° to 60°. How far did he walk?
Competition practiceTrigonometry
A20√3 m
B19 m
C38√3 m
D19√3 m
Step-by-Step Solution
Height above his eyes =28.5 m. Walk =28.5(cot30∘−cot60∘)=28.5⋅32=193 m.
Answer: D) 19√3 m
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