An observer 1.6 m tall is 20 √(3) m away from a tower. The angle of elevation from his eye to the top of the… · ABC26GN5435
Subject: General Aptitude · Chapter: Heights and Distances
An observer 1.6 m tall is $20 \sqrt{3} \mathrm{m}$ away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is :
(a)21.6 m
(b)23.2 m
(c)24.72 m
(d)None of these
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Height above eye = 20√3 × tan 30° = 20 m; tower = 20 + 1.6 = 21.6 m.