From the top of the hundred metre high building than that of depression of a car on the road is 30° then find the distance of the car from the building
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AB- building 100m
angle of dep. 30'
Car at C
angle ACB +30' (alternate angles)
tan 30 +1/root(3) = AB/BC= 100/BC
BC =100*root(3)
angle of dep. 30'
Car at C
angle ACB +30' (alternate angles)
tan 30 +1/root(3) = AB/BC= 100/BC
BC =100*root(3)
Shaurya05:
thank-you brother
np
Answered by
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Given here, Angle of depression=30 degrees.
Height of tower = 100 m
It will make a triangle if you take an point to represent car.
Rotate your figure to left side.
30 degrees will shift to down & 100m will be base.
let the distance between building and car be x m.
therefore
tan 30 degrees=x m/100m
Simply it, you will get 57.74 m as the answer.
Height of tower = 100 m
It will make a triangle if you take an point to represent car.
Rotate your figure to left side.
30 degrees will shift to down & 100m will be base.
let the distance between building and car be x m.
therefore
tan 30 degrees=x m/100m
Simply it, you will get 57.74 m as the answer.
thank-you brother
You are welcome, Hope this helped you.......
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