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?
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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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