Question

A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of ${30}^{\circ }$. A girl standing on the roof of a 20-m-high building finds the angle of elevation of the same bird to be ${45}^{\circ }$. The boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl. [Given $\sqrt{2}=1.41$] [4 MARKS]

Answer 1

Steps: 3 Marks
Answer: 1 Mark

Let O be the position of the bird, B be the position of the boy and FG be the building at which G is the position of the girl.



Let $OL\perp BF$ and $GM\perp OL$. Then,

$BO=100m,\angle OBL={30}^{\circ }$,

$FG=20m$ and $\angle OGM={45}^{\circ }$

From right $\Delta OLB$, we have

$\frac{OL}{BO}=sin{30}^{\circ }\Rightarrow \frac{OL}{100m}=\frac{1}{2}$

$\Rightarrow OL=100m\times \frac{1}{2}=50m$

OM = OL - ML = OL - FG = 50 m - 20 m = 30 m

From right $\Delta OMG$, we have

$\frac{OM}{OG}=sin{45}^{\circ }=\frac{1}{\sqrt{2}}\Rightarrow OG=\sqrt{2}\times OM=\sqrt{2}\times 30m$

$\Rightarrow OG=30\times 1.41m=42.3m$

The distance of the bird from the girl = 42.3 m.