Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
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
\( PQ \ = \ PR \ - \ QR \ \)
\( = \ \sqrt{3}AR \ - \ \frac{AR}{ \sqrt{3}} \)
\( = \ \frac{(3-1)AR}{ \sqrt{3}} \ \)
\( = \ \frac{2 \sqrt{3}}{3} AR \)
\( = \ \frac{2 \sqrt{3}}{3} \ × \ 28.5 \ = \ 19 \sqrt{3} \)
[ \(\because AR \ = \ 30 \ - \ 1.5 \ = \ 28.5 \)m ]
\(\therefore \) the distance walked by the man towards the building is \( 19 \sqrt{3} \) metres.