Question

Two towers AB and CD are of equal height. At a point between them in the line AC joining their bases, the angle of elevation of the nearer tower was observed to be 60o. Then at 24 m from thesame point in a direction perpendicular to AC, the angle of elevation of the top of thetowers are 45ofor the nearer tower and 30ofor the other. Find the height of the towers (h) and the distance apart?

Answer 1

Towers of equal height , nearer tower having elevation of 60° ,  after 24 m , Angle of elevation = 45° and to other tower = 30° Then Height of tower = 36 + 12√3 m & Distance 48√3 + 72   m

Step-by-step explanation:

Let the point be P

Tan 60  =  AB/AP

=> AB = AP √3

After going 24 m then point  M

AM = AP + 24

Tan 45  = AB/AM

=> 1 = AB/(AP + 24)

=> AB = AP + 24

=> AP + 24 = AP√3

=> AP (√3 - 1) = 24

=> AP = 12(√3 + 1)

AB = AP √3  = 12√3(√3 + 1)

CD = AB =  12√3(√3 + 1)

Height of tower = 36 + 12√3 m

Tan30  = CD/CM

1/√3 = 12√3(√3 + 1)/CM

=> CM = 36(√3 + 1)

AC = AM + CM   = AP + 24 + CM

AC = 12(√3 + 1) + 24  + 36(√3 + 1)

=> AC = 48(√3 + 1)  + 24

=> AC = 48√3 + 72

Distance between towers = 48√3 + 72   m

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