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Answer :
Let AB be the building of height h and AC, the horizontal ground through C, the foot of the building. Since the building subtends an angle of 60º at C, hence \( ∠ \ ACB \ = \ 30º \). Let CD be the tower of height 50 m such that \( ∠ \ CAB \ = \ 60º \).
\( ∆ \ BAC \) and \( DCA \) , we have
\( \frac{AC}{AB} \ = \ cot30º \ \)
\( => \ \frac{AC}{h} \ = \ \sqrt{3} \)
\( => \ AC \ = \ \sqrt{3} \) (1)
and \( \frac{DC}{AC} \ = \ tan60º \ \)
\( => \ \frac{50}{AC} \ = \ \sqrt{3} \)
\(=> \ AC \ = \ \frac{50}{ \sqrt{3}} \) (2)
Equating the values of AC from (1) and (2), we get
\( \sqrt{3}h \ = \ 50 \sqrt{3} \ \)
\( => \ h \ = \ \frac{50}{ \sqrt{3}} × \frac{1}{ \sqrt{3}} \)
\( = \ \frac{50}{3} \ = \ 16.66 \)
\(\therefore \) the height of the building is 16.66 m.