A bird is sitting on the top of a vertical pole $$20m$$ high and its elevation from a point O on the ground is $$450$$. It flies off horizontally straight away from the point O. After one second, the elevation of bird from O is reduced to $$300$$, then the speed (in$$ $$m/s), of the bird is

Question

Infinity Learn · Accepted Answer

Let $$AB=$$ height of vertical pole $$=20$$ From ΔOAB,$$tan$$⁡$$45$$∘$$=ABOA$$⇒$$OA=AB=20$$ Let C be the new position of bird  Let $$BC=x$$ From ΔOCD,$$tan$$⁡$$30$$∘$$=20x+20$$⇒$$x=20(3$$−$$1)=14.64$$ $$m/s$$

Applications of trigonometry

Remember concepts with our Masterclasses.

A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45⁰. It flies off horizontally straight away from the point O. After one second, the elevation of bird from O is reduced to 30⁰, then the speed (in m/s), of the bird is

$Let AB = height of vertical pole = 20$
$From ΔOAB, \tan 45^{\circ} = \frac{AB}{OA}$
$\Rightarrow OA = AB = 20$
$\begin{aligned} Let C be the new position of bird \\ Let BC = x \end{aligned}$
$From ΔOCD, \tan 30^{\circ} = \frac{20}{x + 20}$
$\Rightarrow x = 20 (\sqrt{3} - 1) = 14.64 m / s$

Ready to Test Your Skills?

free study material