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 = 600
Angle of depression from Q to A = 450
Here
∠QAR = 450 which is the alternate angle
PQ = 5 m
Take QR = h m
PQ = 5 + h
In right triangle QAR
tan θ = QR/AR
Substituting the values
tan 450 = h/AR
So we get
1 = h/AR
AR = h
In right triangle PAR
tan 600 = PR/AR
Substituting 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.