Question

A 2 m tall girl is standing at some distance from a building of height 20 m. The angle of elevation from her eyes to the top of the building increases from 45º to 60º as she walks towards the building. Then, the distance between the initial position and the final position of the girl is $(\sqrt{3}=1.732)$

• A. 760 cm
• B. 760.8 cm
• C. 706.8 cm
• D. 1040.8 cm

Answer 1

The correct option is B 760.8 cm

Let AB and CD be the girl and the building respectively.
Given that the girl moves towards the building from point A to E.
Let AE = x.
∴ CF = CD – FD
= (20 – 2) m = 18 m …..(i)
$In\Delta CFE,$
$\tan {60}^{\circ }=\frac{CF}{EF}$
$\Rightarrow \sqrt{3}=\frac{18}{EF}$
$\Rightarrow EF=\frac{18}{\sqrt{3}}=6\sqrt{3}m$ .....(ii)
$In\Delta CFA,$
$\tan {45}^{\circ }=\frac{CF}{AF}$
$\Rightarrow 1=\frac{CF}{AF}$
$\Rightarrow CF=AF$
$\Rightarrow CF=AE+EF$
$\Rightarrow 18=x+6\sqrt{3}$ [From (i) and (ii)]
$\Rightarrow x=18-6\sqrt{3}=7.608m=760.8cm$
Hence, the correct answer is option (b).