Question

A man at top of a vertical observation tower observes a car moving at uniform speed coming directly towards the tower. If it takes 12 minutes for angle of depression to change from 30degree to 45 degree, how much after this will the car reach the observation tower?

Answer 1

See in the attachment..........................................

Answer 2

Answer:

It takes 16.43 minutes for the car to reach the observation tower.

Solution:

Let the tower’s height AB be h meters

We know that

Distance=Speed × Time  

Time: it is an on-going sequence of event taking place. Time will be measured in seconds or minutes here as it involves travel.

Speed: It is a distance travelled per unit of time. It measures always in km/h.

Distance: Distance between two points is measured in meters and kilometres.

Distance=v×t

Here  

Speed=v

Time=t

Distance=vt

In Triangle ABC

$\begin{array} { l } { \tan 45 ^ { \circ } = \frac { \mathrm { AB } } { \mathrm { BC } } } \\\\ { \tan 45 ^ { \circ } = \frac { \mathrm { h } } { \mathrm { vt } } } \\\\ { \tan 45 ^ { \circ } = 1 } \\\\ { 1 = \frac { \mathrm { h } } { \mathrm { vt } } } \\\\ { \mathrm { h } = \mathrm { v } \mathrm { t } } \end{array}$

In Triangle ABD

$\begin{array} { l } { \tan 30 ^ { \circ } = \frac { \mathrm { h } } { 12 \mathrm { v } + \mathrm { vt } } } \\\\ { \tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } } } \\\\ { \frac { 1 } { \sqrt { 3 } } = \frac { \mathrm { h } } { 12 \mathrm { v } + \mathrm { vt } } } \\\\ { 12 \mathrm { v } + \mathrm { vt } = \sqrt { 3 } \mathrm { h } } \\\\ { \mathrm { h } = \mathrm { vt } } \end{array}$

$\begin{array} { l } { 12 \mathrm { v } + \mathrm { vt } = \sqrt { 3 } \mathrm { vt } } \\\\ { 12 \mathrm { v } = \sqrt { 3 } \mathrm { vt } - \mathrm { vt } } \\\\ { 12 \mathrm { v } = \mathrm { vt } ( \sqrt { 3 } - 1 ) } \\\\ { \frac { 12 \mathrm { v } } { \mathrm { vt } } = \sqrt { 3 } - 1 } \\\\ { \frac { 12 } { \mathrm { t } } = 1.732 - 1 } \\\\ { \frac { 12 } { \mathrm { t } } = 0.732 } \end{array}$

$\begin{array} { l } { \frac { 1200 } { t } = 73 } \\\\ { t = \frac { 1200 } { 73 } } \\\\ { t = 16.43 \text { Minutes } } \end{array}$