a boy 1.6m tall is 20m away from a tower and observes that the angle of elevation of the top of the tower is 60 degree. Find the height of the tower.
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Answered by
48
let AC be tower
DE=1.6m as BC =1.6 m
DC =20 m as BE = 20 m
so, in triangle ABE
tan 60= AB/BE
√3 =AB/20
AB =20√3m
so,the height of the tower = AC =AB+BC =20√3+1.6
=20×1.732+1.6 =34.64 +1.6
=36.24 m
DE=1.6m as BC =1.6 m
DC =20 m as BE = 20 m
so, in triangle ABE
tan 60= AB/BE
√3 =AB/20
AB =20√3m
so,the height of the tower = AC =AB+BC =20√3+1.6
=20×1.732+1.6 =34.64 +1.6
=36.24 m
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Answered by
45
see the link for the figure of the question
let AB be the height of tower and CD be the boy
we have CD = 20m and CD = 1.6m
also BD = CE = 20m
in triangle ACE ,
tan60 = P/B = AE/20
√3 = AE/20
so, AE = 20√3m
also CD = BE = 1.6m
Height of tower = AE+BE
=(20√3+1.6)m
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