Question

A bird is perched on the top of a tree $30m$ high ans its elevation from a point on the ground is ${45}^o$. It files off horizontally straight away from the observer and in $2$ seconds the elevation of the bird is reduced to ${30}^o$ The speed of the bird is $\left(inm/s\right)$

• A. $14.64$
• B. $21.96$
• C. $10.98$
• D. $12$

Answer 1

The correct option is C $10.98$

Let AB be the tree, where A is the top of the tree and B is its base.

Let the fixed point on the ground be O.

So,

$OB=AB=30mtr.$

Let the bird fly away x mtr from point A to point C.

Draw CD perpendicular to the ground,

So,

CD=AB=30 mtr.

And DB=A=x mtr.

Given elevation angle is ${30}^o$.

In $\Delta COD$

$\tan {30}^o=\frac{CD}{OD}=\frac{CD}{OB+BD}=\frac{30}{30+x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{30}{30+x}$

$\Rightarrow x=30(\sqrt{3}-1)$

$\Rightarrow x=30(1.732-1)$

$\Rightarrow x=30\times \left(0.732\right)$

$\Rightarrow x=21.96$

Now, the bird flies $21.96mtr$. In $2$ second

So,

The speed of the bird is $\frac{21.96}{2}=10.98$ mtr.

Hence, this is the answer.