What is the angle of elevation of the sun when the length of shadow of a vertical pole is equal to its height?
VerifiedHint: Since the length of shadow of vertical pole is equal to the height of the pole. So then angle be tan$\tan \theta = \dfrac{{{\text{height of the pole}}}}{{{\text{length of the shadow of pole}}}}$ as it will form a triangle. On putting the values in this formula, you’ll get the answer.
Complete step-by-step answer:
Given, the length of the shadow of a vertical pole=height of the pole. We have to find the angle of elevation of the sun.
Let’s draw a triangle ABC which has following components-Let AB be the height of the pole, BC be the length of the shadow of the pole on the ground and $\theta $ be the angle of elevation of the sun. Here,$\angle {\text{B = }}{90^ \circ }$ makes the triangle a right angled triangle.
In ∆ABC, $\tan \theta = \dfrac{{\text{P}}}{{\text{B}}}$ where P is perpendicular and B is base of triangle then on putting the values we get,
$ \Rightarrow \tan \theta = \dfrac{{{\text{AB}}}}{{{\text{BC}}}}$
But according to the question, the length of pole AB=the shadow of the pole BC. On putting this value in the formula, we get-
$ \Rightarrow \tan \theta = \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = 1$
We know that $\tan {45^ \circ } = 1$. On putting this in the above formula, we get-
$ \Rightarrow \tan \theta = \tan {45^ \circ } \Rightarrow \theta = {45^ \circ }$
Hence the angle of elevation of sun is ${45^ \circ }$.
Note: We can also solve this question by assuming the length of the pole to be x m and the height of the pole to be h m. so the formula will become -$\tan \theta = \dfrac{{{\text{height of the pole}}}}{{{\text{length of the shadow of pole}}}}$
On putting assumed values, we get-
$ \Rightarrow \tan \theta = \dfrac{{\text{h}}}{{\text{x}}}$
Then it is given that the height of the pole is equal to the length of the pole so h=x. On putting this value-
$ \Rightarrow \tan \theta = \dfrac{{\text{h}}}{{\text{h}}} = 1$
Now put the value of $1 = \tan {45^ \circ }$ in the given eq.-
$ \Rightarrow \tan \theta = \tan {45^ \circ } \Rightarrow \theta = {\tan ^{ - 1}}\left( {\tan {{45}^ \circ }} \right) = {45^ \circ }$
The answer will be the same.
