VerifiedHint: In order to solve this question, we just require the knowledge of a few topics and a good visualization power. By visualizing the given situation, we will draw a diagram and then we can find the height of the lamp post by using the trigonometric ratios and properties of similar triangles.
Complete step-by-step answer:
In the given question, we are asked to find the height of a lamp post which is at a distance of 3.2 m from a given height of 1.6 m and casts a shadow of 4.8 m of the girl on the ground. So, first, let us draw the figure for the given situation. So, the diagram is given below.
In the figure above, we have considered the position of the lamp as point D and the height of the lamp as CD. Also, we have considered the position of the girl at point A and the height of the girl as 1.6 m. Now, we have been given that the girl is at a distance of 3.2 m from the lamp. So, we can say that, AD = 3.2 m. Also, we have been given that the lamp casts a shadow of the girl of length 4.8 m, so we can say that AE = 4.8 m. Now, let us consider each of the conditions given in the question.
i) Trigonometric ratios
Now, according to the conditions given in the question, we can say that triangle BAE and triangle CDE are right angled triangles at angle A and angle D. Now, let us consider $\angle E=\alpha $, which is common in both the triangles under consideration. And we know that the tangent ratio of trigonometry is defined as the ratio of perpendicular to the base. So, we can write,
For, $\Delta BAE,\tan \angle BEA=\dfrac{AB}{AE}$
We have considered $\angle BEA=\alpha $, so we can write it as,
$\tan \alpha =\dfrac{AB}{AE}\ldots \ldots \ldots \left( i \right)$
Now, for $\Delta CDE,\tan \angle DEC=\dfrac{CD}{DE}$
And we can write $\angle DEC=\alpha $, so we get,
$\tan \alpha =\dfrac{CD}{DE}$
Now, according to the figure, we can see that, $DE=AE+AD$, so by substituting it in the above equation, we get,
$\tan \alpha =\dfrac{CD}{AE+AD}\ldots \ldots \ldots \left( ii \right)$
From equation (i) and (ii), we can conclude that,
$\dfrac{AB}{AE}=\dfrac{CD}{AE+AD}$
Now, we will substitute the values of AB, AE, AD which are AB = 1.6 m, AE = 4.8 m and AD = 3.2 m. So, we get,
$\begin{align}
& \dfrac{1.6}{4.8}=\dfrac{CD}{4.8+3.2} \\
& \Rightarrow \dfrac{1.6}{4.8}=\dfrac{CD}{8.0} \\
& \Rightarrow CD=\dfrac{\left( 1.6 \right)\left( 8.0 \right)}{\left( 4.8 \right)} \\
& \Rightarrow CD=2.67 \\
\end{align}$
Hence, we can say that the height of the lamp post is 2.67 m.
ii) Properties of similar triangles.
Now, let us consider triangle BAE and triangle CDE. From the given conditions in the question, we can say that $\angle A=\angle D=90{}^\circ $ and therefore, we can say that AB and CD are parallel. Now, we know that two parallels intersected by a common transversal has a property of corresponding angles and so, we can say angle ABE and angle DCE are equal because they are formed by the intersection of the common transversal CE on the parallel lines AB and CD respectively. And we know that angle BEA and angle CED are equal. Therefore, by the rule of AAA, we can say $\Delta BAE\sim \Delta CDE$. So,
$\dfrac{AB}{CD}=\dfrac{AE}{DE}$
Now, we know that $DE=AE+AD$, so we can write the above equality as,
$\dfrac{AB}{CD}=\dfrac{AE}{AE+AD}$
WE know that, AB = 1.6 m, AE = 4.8 m and AD = 3.2 m. So, by substituting these values, we get,
$\begin{align}
& \dfrac{1.6}{CD}=\dfrac{4.8}{4.8+3.2} \\
& \Rightarrow \dfrac{1.6}{CD}=\dfrac{4.8}{8.0} \\
& \Rightarrow CD=\dfrac{\left( 1.6 \right)\left( 8.0 \right)}{\left( 4.8 \right)} \\
& \Rightarrow CD=2.67 \\
\end{align}$
Hence, we can say that the height of the lamp post is 2.67 m.
Note: The possible mistakes that can be made is in considering the given situation. The shadow of 4.8 m cast by the lamp post can be considered as the distance between the lamp post and the shadow, which will give a wrong answer. Also, remember that the part (i), that is the tangent ratio is the best way to find the height of the lamp post.
