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# The angle of elevation of the top of a tower at a distance $500$ meters from the foot is ${30^\circ }$. The height of the tower is-A.$250\sqrt 3$ metersB.$\dfrac{{500}}{{\sqrt 3 }}$ metersC.$500\sqrt 3$ metersD.$250$ meters

Last updated date: 13th Sep 2024
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Hint: Assume the height of the tower to be AB= h. Then we can use the formula of trigonometric ratio-
$\Rightarrow \tan \theta = \dfrac{P}{B}$ where P is the perpendicular and B is the base of the triangle. Put the value of $\theta = {30^ \circ }$, P=AB and B=BC. Now, we know that $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ . Put this value and the value of AB and BC in the formula and solve it for h.

Given the angle of elevation of the top of the tower at any point in ground$\theta = {30^\circ }$
The distance of the foot of the tower from that point in ground is =$500$ meters
We have to find the height of the tower.
Let us assume that the tower is AB and let the point on the ground be C. And let the height of the tower be h meter.

Here, $\angle ACB = {30^ \circ }$ and $\angle ABC = {90^ \circ }$
AB=h and BC=$500$ meters
Now in $\vartriangle ABC$,
$\Rightarrow \tan \theta = \dfrac{{AB}}{{BC}}$ as AB is the perpendicular here and BC is the base.
On putting the given values we get-
$\Rightarrow \tan {30^ \circ } = \dfrac{h}{{500}}$
Now we know that the value of $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Then on putting this value, we get-
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{{500}}$
On cross multiplication, we get-
$\Rightarrow 500 = h\sqrt 3$
We can also write it as-
$\Rightarrow h\sqrt 3 = 500$
On solving for h, we get-
$\Rightarrow h = \dfrac{{500}}{{\sqrt 3 }}$ meters
Hence the height of the tower is$\dfrac{{500}}{{\sqrt 3 }}$ meters
$\Rightarrow \sin \theta = \dfrac{P}{H}$ and $\cos \theta = \dfrac{B}{H}$ where P is perpendicular, B is base and H is hypotenuse of the triangle.