1) A bridge is situated at right angle to the bank of the river. If one moves away a certain distance from the bridge along this side of the river, the other end of the Bridge is seen at an angle of 45° and if someone moves a further distance of 400 metres in the same direction, the other end is seen at an angle of 30°. Let us find the length of the bridge.
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Answer:
Given :-
To Find :-
Solution :-
Suppose, the length of the bridge AB = x m.
In ∆ ABC,
➛ tan 45° =![\dfrac{AB}{BC} \dfrac{AB}{BC}](https://tex.z-dn.net/?f=%5Cdfrac%7BAB%7D%7BBC%7D)
➛ 1 =![\dfrac{x}{BC} \dfrac{x}{BC}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%7D%7BBC%7D)
➡ BC = x ..... (1)
Again,
⇒ tan 30° =![\dfrac{AB}{BD} \dfrac{AB}{BD}](https://tex.z-dn.net/?f=%5Cdfrac%7BAB%7D%7BBD%7D)
⇒![\dfrac{AB}{BC+ 400} =\: \dfrac{1}{\sqrt{3}} \dfrac{AB}{BC+ 400} =\: \dfrac{1}{\sqrt{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7BAB%7D%7BBC%2B%20400%7D%20%3D%5C%3A%20%5Cdfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D)
➦ BC = AB√3 - 400 .... (2)
From (1) and (2) we get,
Let, x = AB√3 - 400
➟ x - x√3 = - 400
➟ x√3 - x = 400
➟ x(√3 - 1) = 400
➟ x =![\dfrac{400}{\sqrt{3} - 1} \dfrac{400}{\sqrt{3} - 1}](https://tex.z-dn.net/?f=%5Cdfrac%7B400%7D%7B%5Csqrt%7B3%7D%20-%201%7D)
➟ x =![\dfrac{400(√3 + 1)}{(√3 - 1)(√3 + 1)} \dfrac{400(√3 + 1)}{(√3 - 1)(√3 + 1)}](https://tex.z-dn.net/?f=%5Cdfrac%7B400%28%E2%88%9A3%20%2B%201%29%7D%7B%28%E2%88%9A3%20-%201%29%28%E2%88%9A3%20%2B%201%29%7D)
➟ x =![\dfrac{400(√3 - 1)}{3 - 1} \dfrac{400(√3 - 1)}{3 - 1}](https://tex.z-dn.net/?f=%5Cdfrac%7B400%28%E2%88%9A3%20-%201%29%7D%7B3%20-%201%7D)
➟ x =![\dfrac{400(√3 - 1)}{2} \dfrac{400(√3 - 1)}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B400%28%E2%88%9A3%20-%201%29%7D%7B2%7D)
➟ x = 200(1.732 + 1)
➟ x = 200 × 2.732
➠ x = 546.4
Answer:
The length of the bridge 546.4 m