The angles of elevation of top and bottom of a flag kept on a flagpost from 30 metres distance, are 45o and 30o respectively. Height of the flag is [taking $$\sqrt3 = 1.732$$]

Given : CD = 30 m and $$\angle$$ CDB = $$30^\circ$$ and $$\angle$$ CDA = $$45^\circ$$

To find : AB is the flag = $$h$$ = ?

Solution : In $$\triangle$$ BCD,

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

=> $$\frac{1}{\sqrt{3}}=\frac{BC}{30}$$

=> $$BC = \frac{30}{\sqrt{3}}=10\sqrt{3}$$ m

=> $$BC = 10 \times 1.732 = 17.32$$ m

Similarly, in $$\triangle$$ ACD,

=> $$tan(45^\circ)=\frac{AC}{CD}$$

=> $$1=\frac{AC}{30}$$

=> $$AC = 30$$

=> $$AB+BC=30$$

=> $$h=30-17.32=12.68$$ m

=> Ans - (D)