sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

this question has been asked in previous year exams

it is given that A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60 and the angle of depression of the point A from the top of the tower is 45. Find the height of the tower. (Take √3 = 1.732)

question no 33 , heights and distances , ICSE board, ML Aggarwal

Consider QR as the tower

PQ as the pole on it

Angle of elevation from P to a point A is ∠PAR = 60

^{0}Angle of depression from Q to A = 45

^{0}Here

∠QAR = 45

^{0}which is the alternate anglePQ = 5 m

Take QR = h m

PQ = 5 + h

In right triangle QAR

tan θ = QR/AR

Substituting the values

tan 45

^{0}= h/ARSo we get

1 = h/AR

AR = h

In right triangle PAR

tan 60

^{0}= PR/ARSubstituting the values

√3 = (5 + h)/ h

So we get

√3h = 5 + h

h (√3 – 1) = 5

h (1.732 – 1) = 5

By further calculation

0.732 h = 5

h = 5/0.732 = 5000/732

h = 6.83

Hence, the height of the tower is 6.83 m.