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Q6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30º to 60º as he walks towards the building. Find the distance he walked towards the building.
Answer :


Let OA be the building and PL be the initial position of the man such that \( ∠ \ APR=30º \) and AO = 30 m. Let MQ be the position of the man at a distance PQ. Here \( ∠ \ AQR \ = \ 60º \).



Now from \( ∆ \ ARQ \) and \( ∆ \ ARP \) , we have

\( \frac{QR}{AR} \ = \ cot60º \ => \ \frac{QR}{AR} \ = \ \frac{1}{ \sqrt{3}} \ => \ QR \ = \ \frac{AR}{ \sqrt{3}} \)      (1)

and, \( \frac{PR}{AR} \ = \ cot30º \ => \ \frac{PR}{AR} \ = \ \sqrt{3} \ => \ PR \ = \ \sqrt{3}AR \)      (2)

From (1) and (2), we get

\( PQ \ = \ PR \ - \ QR \ = \ \sqrt{3}AR \ - \ \frac{AR}{ \sqrt{3}} \) \( = \ \frac{(3-1)AR}{ \sqrt{3}} \ = \ \frac{2 \sqrt{3}}{3} AR \)

\( = \ \frac{2 \sqrt{3}}{3} \ × \ 28.5 \ = \ 19 \sqrt{3} \)      [∵ \( AR \ = \ 30 \ - \ 1.5 \ = \ 28.5 \)m ]

∴ the distance walked by the man towards the building is \( 19 \sqrt{3} \) metres.