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# A girl of height $1.8m$ is walking away from the base of a lamp post at a speed of $1.2m{s^{ - 1}}$. If the lamp is $4.5m$ above the ground, find the length of her shadow after $5$ seconds.

Last updated date: 20th Jun 2024
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Hint: First we have to calculate the distance of the girl from the lamp post after $5$ seconds. Then express the given data in a figure. Then applying the properties of right-angled triangles and trigonometric equations, we get the desired result.

Useful formula:
If the speed of a body is $x$ metre per second, then the distance covered by the body in $t$ seconds is speed $\times$ time $= xt$.
In a right angled triangle $ABC$ with ${90^ \circ }$ at $A$ and one of the non-right angles, say $\angle B = \theta$ , then $\tan \theta = \dfrac{{AC}}{{AB}}(\dfrac{{{\text{Opposite}}}}{{adjacent}})$

Complete step by step solution:
Given, the height of the girl is $1.8m$.
The speed of the girl is $1.2m{s^{ - 1}}$.
Height of the lamp is $4.5m$.
We have to find the length of her shadow after $5$ seconds.
Consider the figure.

Let $PQ$ represent the lamp post, $PQ = 4.5m$.
$A$ represents the position of the girl after $5$ seconds.
$\Rightarrow AB = 1.8m$
It is given that the girl travels at a speed of $1.2m{s^{ - 1}}$.
Distance covered by her is speed $\times$ time.
Means she covers $1.2 \times 5 = 6m$ in $5$ seconds.
This gives $AP = 6m$.
Draw a line parallel to $AP$ passing through $B$.
Join $BQ$ and extend to meet the line $AP$ at $C$.
Then since $CP$ parallel to $BR$ and $CQ$ is a common line intersecting these lines we have,
$\angle C = \angle B = \theta$
Consider $\vartriangle BRQ$, $BR = AP = 6m$ ( since $AB = PR$)
$\therefore \tan \theta = \dfrac{{{\text{Opposite}}}}{{{\text{adjacent}}}} = \dfrac{{QR}}{{BR}} = \dfrac{{2.7}}{6} - - - (i)$
Now consider $\Delta CAB$, here $AB = 1.8m$
$\therefore \tan \theta = \dfrac{{{\text{Opposite}}}}{{{\text{adjacent}}}} = \dfrac{{AB}}{{AC}} = \dfrac{{1.8}}{{AC}} - - - (ii)$
According to the figure $AC$ is the shadow of the girl after $5$ seconds.
From $(i)$ and $(ii)$ we have, $\dfrac{{2.7}}{6} = \dfrac{{1.8}}{{AC}}$
$\Rightarrow AC = \dfrac{{(1.8) \times 6}}{{2.7}} = \dfrac{{2 \times 6}}{3} = 4$.

Therefore, the length of the shadow of the girl after $5$ seconds is $4m$.

In a right-angled triangle with one of the non-right angles $\theta$, then,
$\sin \theta = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$
$\cos \theta = \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}$
$\tan \theta = \dfrac{{{\text{Opposite}}}}{{{\text{adjacent}}}}$