Ex.9.1 Q15 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of \(30^\circ\), which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be \(60^\circ.\) Find the time taken by the car to reach the foot of the tower from this point.

   

Text Solution

  

What is Known?

(i) Angle of depression is \(30^\circ\)

(ii) \(6\) seconds later angle of depression is \(60^\circ\)

What is Unknown?

Time taken by the car to reach the foot of the tower \(= CD\)

Reasoning:

Let the height of the tower as \(AD\) and the starting point of the car as \(B\) and after \(6\) seconds point of the car as \(C\). The angles of depression of the car from the top \(A\) of the tower at point \(B\) and \(C\) are \(30^\circ\) and \(60^\circ\) respectively.

Distance travelled by the car from the starting point towards the tower in \(6\) seconds \(= BC\)

Distance travelled by the car after 6 seconds towards the tower \(= CD\)

Trigonometric ratio involving \(AD, \,BC, \,CD\) and angles is \(tan\,\theta \).

We know that,

\[\text {Speed} = \frac { \text { Distance } } { \text { Time } }\]

The speed of the car is calculated using the distance \(BC\) and time \(= 6\,\rm seconds.\) seconds. Using Speed and Distance \(CD,\) time to reach foot can be calculated.

Steps:

In \(\Delta ABD\),

\[\begin{align}\tan {{30}^{0}}&=\frac{AD}{BD} \\ \frac{1}{\sqrt{3}}&=\frac{AD}{BD} \\BD&=AD\sqrt{3}\,\,\,....(1)  \end{align}\]

In \(\Delta ACD\),

\[\begin{align} \tan {{60}^{0}}&=\frac{AD}{CD} \\  \sqrt{3}&=\frac{AD}{CD} \\  AD&=CD\sqrt{3}\,\,\,....(2)\text{ }  \end{align}\]

From equation (1) and (2)

\[\begin{align}BD&=CD\sqrt{3}\times \sqrt{3} \\ BC+CD&=3CD\qquad \qquad\left[ \because BD=BC+CD \right] \\ BC&=2CD\,\,\,....(3)\text{ }  \end{align}\]

Distance travelled by the car from the starting point towards the tower in \(6\) seconds \(= BC\) Speed of the car to cover distance \(BC\) in \(6\) seconds;

\[\begin{align}\text Speed&=\,\frac{\text{Distance}}{\text{Time}} \\ & =\,\frac{BC}{6\,} \\ &=\,\frac{2CD}{6}\,\qquad\left[ \text{from(3)} \right] \\ & =\,\frac{CD}{3}\, \end{align}\]

Speed of the car \(=\,\begin{align}\frac{CD}{3}\,\rm{m/s}\end{align}\)

Distance travelled by the car from point \(C\), towards the tower \(= CD\)

Time to cover distance \(CD\) at the speed of \(=\,\begin{align}\frac{CD}{3}\,\rm{m/s}\end{align}\)

\[\begin{align}\text Time   &=\frac{\text{Distance}}{\text{Speed}} \\ & =\frac{CD}{\frac{CD}{3}}\, \\ & =CD\times \frac{3}{CD} \\ & =3  \end{align}\]

Time taken by the car to reach the foot of the tower from point \(C\) is \(3\) seconds.