anyone can just help me out with dat...plz itss urgent..
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Secondary School
Math
8+4 pts
From the top of the building 60m high the angle of depression of the top and bottom of a vertical lamp post are observed to be 30 degree and 60 degree respectively find a)the distance between the building and the lamp post b)the height of the lamp post
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by CuteamanjuAbhirst 07.12.2016
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Geekydude121
Samaritan
Let AB be the building
And CD be the lamp post. While DE is the horizontal line parallel to the ground from the top of the lamp post to the building.
So in Triangle ABC,
AB=60m
θ=60°
tan θ = perpendicular /base
tan 60°= AB / BC
√3=60 / BC
BC = 60 / √3
On rationalising denominator, we get
BC=60 * √3 /3
=[20 * √3]m
Distance between building and lamp post =20 √3 cm
=20*1.732(√3=1.732)
=34.64m(Ans)
EBCD is a rectangle , hence BC =ED
In Triangle AED
θ=30°
tan 30° = AE / ED
1 /√3 = AE / 20√3AE * √3 = 20√3AE = 20mAB = AE + EBEB =60 - 20=40m
Since EB = CD (EBCD being a rectangle)CD or Height of lamp post = 40m (Ans)
Secondary School
Math
8+4 pts
From the top of the building 60m high the angle of depression of the top and bottom of a vertical lamp post are observed to be 30 degree and 60 degree respectively find a)the distance between the building and the lamp post b)the height of the lamp post
Advertisement
Report
by CuteamanjuAbhirst 07.12.2016
Answers


Geekydude121
Samaritan
Let AB be the building
And CD be the lamp post. While DE is the horizontal line parallel to the ground from the top of the lamp post to the building.
So in Triangle ABC,
AB=60m
θ=60°
tan θ = perpendicular /base
tan 60°= AB / BC
√3=60 / BC
BC = 60 / √3
On rationalising denominator, we get
BC=60 * √3 /3
=[20 * √3]m
Distance between building and lamp post =20 √3 cm
=20*1.732(√3=1.732)
=34.64m(Ans)
EBCD is a rectangle , hence BC =ED
In Triangle AED
θ=30°
tan 30° = AE / ED
1 /√3 = AE / 20√3AE * √3 = 20√3AE = 20mAB = AE + EBEB =60 - 20=40m
Since EB = CD (EBCD being a rectangle)CD or Height of lamp post = 40m (Ans)
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