
A vertical pole of $5.6$ meter height casts a shadow 3.2 meter long. At the same time find the
height of the pole which casts a shadow 5 meter long.
Answer
584.4k+ views
Hint: Use the trigonometric ratios to find the relation between the height of the pole and the
length of the shadow and use it to find the desired height of the pole.
It is given that a vertical pole of $5.6$ meter height casts a shadow 3.2 meter long. At the same
time, another pole casts a shadow 5 meters long.
We have to find the height of the other pole.
Assume that the shadow is making an angle$\alpha $ with the top of the pole, then take a look at
the figure of the first pole and its shadow:
Use the trigonometric ratio to find the tan of the angle$\alpha $ , which is the ratio of the
opposite side and the adjacent side of the triangle.
$\tan \alpha = \dfrac{{{\text{opposite side}}}}{{{\text{Adjacent side}}}}$
We know that the opposite side is $AB$ and the adjacent side is $BC$ and the values of these
sides are:
$AB = 5.6$m and $BC = 3.2$m
Substitute these values in the trigonometric ratio:
$\tan \alpha = \dfrac{{AB}}{{BC}}$
$ \Rightarrow \tan \alpha = \dfrac{{5.6}}{{3.2}}$
$ \Rightarrow \tan \alpha = \dfrac{{56}}{{32}}$
$ \Rightarrow \tan \alpha = \dfrac{7}{4}$ … (1)
Now, take a look at the figure of the pole who is casting the shadow 5 meters long and assume that the height of the pole is $x$meters.
Use the trigonometric ratio to find the tan of the angle$\alpha $ , which is the ratio of the
opposite side and the adjacent side of the triangle.
$\tan \alpha = \dfrac{{{\text{opposite side}}}}{{{\text{Adjacent side}}}}$
We know that the opposite side is $A'B'$ and the adjacent side is $B'C'$ and the values of these
sides are:
$A'B' = x$m and $B'C' = 5$m
Substitute these values in the trigonometric ratio:
$\tan \alpha = \dfrac{{A'B'}}{{B'C'}}$
$ \Rightarrow \tan \alpha = \dfrac{x}{5}$
Substitute the value of the $\tan \alpha $ from the equation (1) so we have,
$\dfrac{7}{4} = \dfrac{x}{5}$
$ \Rightarrow x = \dfrac{{35}}{4}$
$ \Rightarrow x = 8.75$
Note: We can also solve this problem by comparing the proportion between the length of the
pole length and the length of the shadow.
length of the shadow and use it to find the desired height of the pole.
It is given that a vertical pole of $5.6$ meter height casts a shadow 3.2 meter long. At the same
time, another pole casts a shadow 5 meters long.
We have to find the height of the other pole.
Assume that the shadow is making an angle$\alpha $ with the top of the pole, then take a look at
the figure of the first pole and its shadow:
Use the trigonometric ratio to find the tan of the angle$\alpha $ , which is the ratio of the
opposite side and the adjacent side of the triangle.
$\tan \alpha = \dfrac{{{\text{opposite side}}}}{{{\text{Adjacent side}}}}$
We know that the opposite side is $AB$ and the adjacent side is $BC$ and the values of these
sides are:
$AB = 5.6$m and $BC = 3.2$m
Substitute these values in the trigonometric ratio:
$\tan \alpha = \dfrac{{AB}}{{BC}}$
$ \Rightarrow \tan \alpha = \dfrac{{5.6}}{{3.2}}$
$ \Rightarrow \tan \alpha = \dfrac{{56}}{{32}}$
$ \Rightarrow \tan \alpha = \dfrac{7}{4}$ … (1)
Now, take a look at the figure of the pole who is casting the shadow 5 meters long and assume that the height of the pole is $x$meters.
Use the trigonometric ratio to find the tan of the angle$\alpha $ , which is the ratio of the
opposite side and the adjacent side of the triangle.
$\tan \alpha = \dfrac{{{\text{opposite side}}}}{{{\text{Adjacent side}}}}$
We know that the opposite side is $A'B'$ and the adjacent side is $B'C'$ and the values of these
sides are:
$A'B' = x$m and $B'C' = 5$m
Substitute these values in the trigonometric ratio:
$\tan \alpha = \dfrac{{A'B'}}{{B'C'}}$
$ \Rightarrow \tan \alpha = \dfrac{x}{5}$
Substitute the value of the $\tan \alpha $ from the equation (1) so we have,
$\dfrac{7}{4} = \dfrac{x}{5}$
$ \Rightarrow x = \dfrac{{35}}{4}$
$ \Rightarrow x = 8.75$
Note: We can also solve this problem by comparing the proportion between the length of the
pole length and the length of the shadow.
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