from a window 60m high above of an orginisation working for consumer protection angle of depression of two cars on the straight road leading to the foot of building are 30degree and 60degree if one car is exactly behind other find distance between two cars (use root 3=1.73)
Answers
Let Car 1 be A and Car 2 be B.
Height of building = CD = 60m .
angle A =30° and Angle B = 60°
Tan 60° =
Rationalising the denominator
BC = 20√3 m
Tan 30° =
AB + BC = 60√3
AB + 20√3 = 60√3
AB =60√3 -20√3
AB =40√3 m
AB =40×1.73
AB =69.2 m
So the distance between two cars is
69.2 meters.
Step-by-step explanation:
Let Car 1 be A and Car 2 be B.
Height of building = CD = 60m .
angle A =30° and Angle B = 60°
Tan 60° = \sf\frac{Side\: opposite\: to \:angle}{Side \:adjacent\: to \:angle}
Sideadjacenttoangle
Sideoppositetoangle
\sqrt{3}=\sf\frac{cd}{bc}
3
=
bc
cd
\sqrt{3} = \frac{60}{bc}
3
=
bc
60
bc = \frac{60}{ \sqrt{3} }bc=
3
60
Rationalising the denominator
bc = \frac{60}{ \sqrt{3} } \times \frac{\sqrt{3}}{ \sqrt{3} }bc=
3
60
×
3
3
bc = \frac{60 \sqrt{3} }{3}bc=
3
60
3
BC = 20√3 m
Tan 30° = \sf\frac{Side \:opposite\: to\: angle}{Side \:adjacent \:to\: angle}
Sideadjacenttoangle
Sideoppositetoangle
\frac{1}{ \sqrt{3} } = \frac{60}{ac}
3
1
=
ac
60
ac = 60 \sqrt{3}ac=60
3
AB + BC = 60√3
AB + 20√3 = 60√3
AB =60√3 -20√3
AB =40√3 m
AB =40×1.73
AB =69.2 m
So the distance between two cars is
69.2 meters.