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A peacock is sitting on the tree and observes its prey on the ground .It makes an angle of depression of 22°to catch the prey. the speed of the peacock was 10km/hr and it catches the prey in 1min 12seconds.At 2hat heught was the peacock on the tree? |
Answer» solution\frac{distance}{time}=speed time distance
=speed \therefore AB=300\frac{m}{min}\times 12 seconds∴AB=300 min m
×12seconds \frac{300}{60}\times 12 60 300
×12 AB = 60 m. Now, tan30^{\circ}=\frac{H}{AB}tan30 ∘ = AB h
\therefore height (h)=TAN(30^{\circ})\times AB∴height(h)=tan(30 ∘ )×AB \frac{1}{\sqrt{3}}\times 60=\frac{60}{\sqrt{3}}metere. 3
1
×60= 3
60
metere. = 34.64 m |
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