Question: A person is standing on a tower of height $15 ( \sqrt { 3 } + 1 ) m$ and observing a car coming ...

Question

A person is standing on a tower of height $15 ( \sqrt { 3 } + 1 ) m$ and

observing a car coming towards the tower. He observed that

angle of depression changes from $30 ^ { \circ }$ to $45 ^ { \circ }$ in 3 sec. What

is the speed of the car

Answer 1

$A B = O P [ \cot \alpha - \cot \beta ]$ , where α = 30^o, β = 45^o

⇒ $= 15 ( 3 - 1 )$ = 30 metre.

Speed = $\frac { 30 } { 3 } = 10 \mathrm {~m} / \mathrm { sec }$

=

= 36 km/hr

Question: A person is standing on a tower of height \(15 ( \sqrt { 3 } + 1 ) m\) and observing a car coming ...