A 1.6 m tall observer is 45 meters away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in meters is (Take √3 = 1.732)

Given : CE is the observer of height = 1.6 m and he is at distance of DE = 45 m from tower

To find : Height of tower AD = ?

Solution : By symmetry, BC = DE = 45 m and BD = CE = 1.6 m

In $$\triangle$$ ABC,

=> $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(30^\circ)=\frac{AB}{45}$$

=> $$\frac{1}{\sqrt{3}}=\frac{AB}{45}$$

=> $$AB=\frac{45}{\sqrt{3}}=15\sqrt{3}$$ m

=> $$AB=15 \times 1.732=25.98$$ m

$$\therefore$$ Height of tower, AD = AB + BD

= $$25.98+1.6 = 27.58$$ m

=> Ans - (C)