ABC is a triangular park with AB = AC = 100 metres.

Question

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If
the angles of elevation of the top of the tower at A and B are cot--1
 (3$$\sqrt 2 $$ ) and cosec--1
(2$$\sqrt 2 $$ ) respectively,
then the height of the tower (in metres) is :

Accepted Answer

$$\Delta $$APM

$${h \over {AM}} = {1 \over {3\sqrt 2 }}$$

$$\Delta $$BPM

$${h \over {BM}} = {1 \over {\sqrt 7 }}$$

$$\Delta $$ABM

AM2 + MB2 = (100)2

'
$$ \Rightarrow $$ 18h2 + 7h2 = 100 \times  100

$$ \Rightarrow $$ h2 = 4 \times  100

$$ \Rightarrow $$ h = 20

Solutions · 1