One observer estimates the angle of elevation to the basket of a hot air balloon to be 60° , while another observer…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 5 Marks · March 2023 · Standard
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465 Marks · March 2023 · Standard
One observer estimates the angle of elevation to the basket of a hot air balloon to be $60^{\circ}$, while another observer $100$ m away estimates the angle of elevation to be $30^{\circ}$. Find : (a) The height of the basket from the ground. (b) The distance of the basket from the first observer's eye. (c) The horizontal distance of the second observer from the basket.
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Correct Figure Let $B$ is the basket of hot air balloon. $\tan 60^{\circ} = \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ quad (i) $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{x + 100} \Rightarrow x = h\sqrt{3} - 100$ quad (ii) using (i) and (ii) (a) $h= (h\sqrt{3}-100)\sqrt{3} = 3h - 100\sqrt{3} \Rightarrow h = 50\sqrt{3}$ m (b) $\sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{h}{y} \Rightarrow y = \frac{50\sqrt{3}}{\sqrt{3}/2} = 100$ m (c) $x = \frac{h}{\sqrt{3}} = 50$ m $\Rightarrow x+100=150$ m # ANOTHER SOLUTION AS PER BELOW FIGURE IS ALSO POSSIBLE Correct Figure Let $B$ is the basket of hot air balloon. $D$ and $C$ be the positions of the first and second observer's respectively. $\tan 60^{\circ} = \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ quad (i) $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{100-x} \Rightarrow \sqrt{3}h = 100 - x$ quad (ii) (a) using (i) and (ii) $h = \sqrt{3} (100 - \sqrt{3}h) \Rightarrow h = 25\sqrt{3}$ m (b) $\sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{h}{BD} \Rightarrow BD = 50$ m (c) $x = \frac{h}{\sqrt{3}} = 25$ m $AC = 100 - x = 75$ m
