The angle elevation of the top Q of a vertical tower PQ from a point Xon the ground is 60 degree . at a point Y, 40 m verticaaly above X, the angle of elevation is 45 degree . find the height of the tower PQ and the distance XY.
Answers
Answer:
id 683 954 0964
Pas 100200
now on zooom ❤️❤️❤️
Solution :-
In right ∆QRY,
→ Tan 45° = QR / RY
→ 1 = x / RY
→ RY = x m .
from image we can see that,
- RP = YX = 40 m.
- RY = PX
In right ∆QPX ,
→ Tan 60° = QP / PX
→ √3 = (QR + RP) / RY
→ √3 = (x + 40)/x
→ √3x = x + 40
→ √3x - x = 40
→ x(√3 - 1) = 40
→ x = 40/(√3 - 1) * {(√3 + 1)/(√3 + 1)}
→ x = 40(√3 + 1)(3 - 1)
→ x = 20(√3 + 1) = Distance PX . (Ans.)
then,
→ PQ = x + 40
→ PQ = 20(√3 + 1) + 40
→ PQ = 20(√3 + 1 + 2)
→ PQ = 20(√3 + 3) m (Ans.)
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