Question

40. (1) Let ( A C ) be the tower ( therefore quad frac{A C}{B C}=tan 45^{circ} Rightarrow B C=A C=30 mathrm{m} )
and ( quad frac{A C}{C D}=tan 60^{circ} )
[
C D=frac{A C}{sqrt{3}}=frac{30}{sqrt{3}}=10 sqrt{3} mathrm{m}
]
So, required distance ( =(30+10 sqrt{3}) mathrm{m} )
( =(30+17.32)=47.32 mathrm{m} )

# 40. The angles of elevation of the top of a tower 30 m. High, from two points on the level ground on its opposite sides are 45° and 60° what is the distance between the two points? (1) 47.32 m (2) 41.23 m (3) 38.12 m (4) 52.10 m

Solution