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A boy standing on a horizontal plane is flying a kite with a string of length $60$ m, at an angle of elevation of $30^{\circ}$. Another boy standing on the roof of a $20$ m high building, finds the angle of elevation of same kite to be $45^{\circ}$. If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $\sqrt{3} = 1.73$)
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Sol. Correct figure (1 Mark)
Let K be the position of kite.
Let KL $= h$, CP $= x$ & PB$= y$
In rt. angled $\triangle KPB$
$\frac{h+20}{60} = \sin 30^{\circ} = \frac{1}{2}$ (1/2 Mark)
$\Rightarrow h = 10$ (1/2 Mark)
Also, $\frac{y}{60} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ (1/2 Mark)
$\Rightarrow y = 30\sqrt{3} = 30 \times 1.73 = 51.9$ (1/2 Mark)
In rt. angled $\triangle KLM$
$\frac{10}{x} = \tan 45^{\circ} = 1$ (1/2 Mark)
$\Rightarrow x = 10$ (1/2 Mark)
Height of kite from ground is $10 + 20 = 30$ m (1/2 Mark)
Distance of first boy from the base of the building $= 51.9 + 10 = 61.9$ m (1/2 Mark)
Let K be the position of kite.
Let KL $= h$, CP $= x$ & PB$= y$
In rt. angled $\triangle KPB$
$\frac{h+20}{60} = \sin 30^{\circ} = \frac{1}{2}$ (1/2 Mark)
$\Rightarrow h = 10$ (1/2 Mark)
Also, $\frac{y}{60} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ (1/2 Mark)
$\Rightarrow y = 30\sqrt{3} = 30 \times 1.73 = 51.9$ (1/2 Mark)
In rt. angled $\triangle KLM$
$\frac{10}{x} = \tan 45^{\circ} = 1$ (1/2 Mark)
$\Rightarrow x = 10$ (1/2 Mark)
Height of kite from ground is $10 + 20 = 30$ m (1/2 Mark)
Distance of first boy from the base of the building $= 51.9 + 10 = 61.9$ m (1/2 Mark)