Question

From the top of 97m high building, the angle of elevation of the top of a tower is 60' and the ang
depression of the foot is 45° find the height of the tower.​

Answer 1

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Answer :-

Height of tower = 97(√3+1)m

Answer 2

$\huge{\boxed{\bold{Question}}}$

From the top of 97m high building, the angle of elevation of the top of a tower is 60' and the ang

depression of the foot is 45° find the height of the tower.

Answer

In ∆DBC

Tan45°=DC/BC

1=97/BC

BC=97---------------(1)

again In ∆ADE

Tan 60°=AE/DE

√3=H/97

97√3=H. [•°• BC=DE=97m]

Now the height of tower =AB

AB=AE+BE

AB=AE+BE

=97√3+97

AB=AE+BE

=97√3+97AB

=97(√3+1)m