Asked by tawheed
PQ is a tower standing vertically on the horizontal ground. from A, the angle of elevation of top P of the tower is 45°. On moving 50 m. up a slope of 15°, the angle of elevation of P is found to be 75° from B. The horizobtal through B is BC. find height of the tower.
Answers
Answered by
Henry
Tan45 = h/d, h = d*Tan45.
Tan75 = h/(d-50), h = (d-50)*Tan75.
Therefore, h = d*Tan45 = (d-50)*Tan75.
d*Tan45 = (d-50)*Tan75.
divide both sides by Tan75:
0.268d = d-50,
d-0.268d = 50, d = 68.3 m.
h = d*Tan45 =
d = The hor. distance between point A and the foot of the tower.
Tan75 = h/(d-50), h = (d-50)*Tan75.
Therefore, h = d*Tan45 = (d-50)*Tan75.
d*Tan45 = (d-50)*Tan75.
divide both sides by Tan75:
0.268d = d-50,
d-0.268d = 50, d = 68.3 m.
h = d*Tan45 =
d = The hor. distance between point A and the foot of the tower.
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