The angle of elevation of the top of a vertical tower from a point A, due east of it is 45…

Question

The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o. The angle of elevation of the top of the same tower from a point B, due south of A is 30o. If the distance between A and B is $$54\sqrt 2 \,m,$$ then the height of the tower (in metres), is :

Accepted Answer

Let the height of tower = h

In triangle PQA,

tan45o = $${h \over {QA}}$$

$$ \Rightarrow $$   h = QA

In triangle  PQB,

tan300 = $${h \over {BQ}}$$

$$ \Rightarrow $$   BQ = $$\sqrt 3 $$h

In triangle BAQ

$$\angle $$ QAB = 90o.

$$ \therefore $$    QA2 + AB2 = QB2

$$ \Rightarrow $$   h2 + (54$$\sqrt 2 $$)2 = ($$\sqrt 3 $$h)2

$$ \Rightarrow $$   2h2 = (54$$\sqrt 2 $$)2

$$ \Rightarrow $$   $$\sqrt 2 h$$ = 54$$\sqrt 2 $$

$$ \Rightarrow $$   h = 54 m

Solutions · 1