Example: Consider a person standing on the top of tower A and the angle of elevation of the top of tower B
at a distance of 18 meters from the first tower A, is 45° and the angle of depression of the bottom
of tower B is 30° . Find the height of the tower A and B.
Height of tower B= OQ+QP
In triangle MQO
tan45° = ${OQ}/{MQ}$
1 = ${OQ}/{18}$
OQ= 18 m
In triangle MQP
tan 35° = ${QP}/{MQ}$
$1/√3$ = ${QP}/{18}$
QP = $18/√3$ = $6√3$ m
Height of tower A = QP = MN = $6√3$ m
Therefore height of tower B = OQ+QP = 18+ 6√3 = 28.38 m
