Question

A parachute is tied to a stone (A). Initially a man was there on parachute when it makes ${60}^{\circ }$ with the horizontal plane. After sometime the man ties the rope to himself and then jumps off from the parachute. At some point C man makes ${30}^{\circ }$ with the horizontal plane. From B, an imaginary line is drawn perpendicular to the ground so that it cuts AC (in the fig) in D. Distance BD is ___ m (Round off to nearest integer)

Answer 1

In the figure,
$cos{60}^{\circ }=\frac{AE}{AB}\Rightarrow AB\times \frac{1}{2}=AE\Rightarrow 10m=AE$
Also,$sin{60}^{\circ }=\frac{BE}{AB}\Rightarrow 20\times \frac{\sqrt{3}}{2}=BE\Rightarrow BE=10\sqrt{3}mtan{30}^{\circ }=\frac{DE}{AE}\Rightarrow DE=AE\times tan{30}^{\circ }=\frac{10}{\sqrt{3}}\therefore BD=BE-DE=10\sqrt{3}-\frac{10}{\sqrt{3}}=10\left(\frac{3-1}{\sqrt{3}}\right)=\frac{20}{\sqrt{3}}m$