Question

The angle of elevation of the top q of a vertical tower pq from point x on the ground is 60 degree.At a point y 40m vertically above x the angle of elevation is 45 degree .Find the height of the tower pq and distance xq.

Answer 1

in triangle PQX ;
perpendicular / base = tan60
PQ / XP = √3
QA + AP / XP = √3
QA + 40 / xp = √3 -----(AP=XY=40)
QA = XP √3 - 40
In triangle QAY ;
perpendicular / base = TAN45
QA / AY = 1
QA = AY
XP √3 - 40 = XP ------(XP = AY)
XP√3 - XP = 40
XP (√3-1) = 40
XP = 40 / (√3-1)
XP= 40 / (√3-1) × (√3+1) / (√3+1) -------(by rationalization)
XP = 40 (√3+1) / (√3)^2 - (1)^2 ----{(A+B) (A-B) = A^2 - B^2 }
XP = 40 ( 1.73 + 1 ) / 3-1
XP = 40 × 2.73 / 2
XP = 109.2 / 2
XP = 54.6 ans
again In triangle PQX
PQ / XP = √3
QA + AP = √3 × XP
QA + 40 = 1.73 × 54.6
QA = 94.46 - 40
QA = 54.46 ans
answers are
QA = 54.46
XP = 54.6
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Answer 2

Step-by-step explanation:

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