There are two temples, one on each bank of a river, just opposite to each other. One temple is 54 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are $$30^\circ$$ and $$60^\circ$$ respectively. Find the width of the river and the difference of heights of the temples.

It is given that Temple AD = 54 m and

Solution : In right $$\triangle$$ ADE,

=> $$tan(60^\circ)=\frac{AD}{DE}$$

=> $$\sqrt3=\frac{54}{DE}$$

=> $$DE=\frac{54}{\sqrt3}$$

=> $$DE=18\sqrt3\approx31.18$$ m width

Similarly, in right $$\triangle$$ ABC,

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

=> $$\frac{1}{\sqrt3}=\frac{AB}{18\sqrt3}$$

=> $$AB=18$$

$$\therefore$$ Height of temple = CE = $$54-18=36$$ m height

Difference of heights=54-36=18