The angle of elevation of the top of a tower from certain point is 30°.If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
(Class 10 Maths Sample Question Paper)
(Class 10 Maths Sample Question Paper)
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FIGURE IS IN THE ATTACHMENTLet the height of the tower PR be h m , the angle of elevation at point Q is 30°(∠PQR = 30°). S be the position of opposite if observer after moving 20 m towards a tower.
ATQ
∠PSR = ∠ PQR + 15 °
∠PSR = 30° + 15 ° = 45°
∠PSR = 45°
In right angled ∆PRS
tan∠PSR =PR / SR = h /x
[ tanϴ = P / B]
tan 45° = h /x
1 = h /x [ tan 45° = 1]
x = h …………………..(1)
In right angled ∆PRQ
tan 30° = PR / QR = PR / (QS + SR)
[ QR= QS + SR]
1/√3 = h / (20 + x)
20 + x = √3h
20 + h = √3h (from eq 1)
20 = √3h - h
20 = h (√3 -1)
h = 20 /(√3-1)
h = 20 × (√3 + 1) /(√3-1) (√3 + 1)
[ By rationalising]
h = 20 (√3 + 1) / (√3² - 1²) [ (a+b) (a-b)=a²-b²]
h = 20 (√3 + 1) / 3 -1
h = 20 (√3 + 1) / 2
h = 10 (√3 + 1) m.
Hence, the required height of the tower is 10 (√3 + 1) m.
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