A castle stands on top of a mountain. At a point on level ground which is 55 m away from the foot of the mountain, the angle of elevation of the top of the castle and the top of the mountain are 60° and 45° respectively. Find the height of the castle.
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Given :- (From diagram)
- BC = Height of mountain = H
- AB = Height of castle = h .
- CD = 55m .
- ∠BDC = 45°
- ∠ADC = 60° .
To Find :-
- Height of castle = h ?
Solution :-
in Right ∆BCD, we have ,
→ Tan45° = BC/CD
→ 1 = H/55
→ H = 55m . --------- Eqn.(1)
in Right ∆ACD , we have,
→ Tan60° = AC/CD
→ √3 = (H + h) / 55
→ √3*55 = H + h
Putting value from Eqn.(1)
→ 55√3 = 55 + h
→ h = 55√3 - 55
→ h = 55(√3 - 1)
→ h = 55(1.732 - 1)
→ h = 55 * 0.732
→ h = 40.26 m .(Ans.)
Hence, the height of the castle is 40.26m.
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