from the top of a 60 m high building AB the angle of depression of top and bottom of a vertical lamp post CD are observed to be 30 and 60 respectively . find horizontal distance between AB and CD?
Answers
Answered by
3
HI MATE
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)
Monica695:
HEY flower 161 can you tell me how can I put the stickers on the I am sad because I put the stickers but I can't see it aur how can I select the writing style please tell me and how I do the maths numerator upon denominator
TO GET BOLD , ITALIC AND UNDERLINED FONT STICKERS HAVE TO USE BRAINLY IN COMPUTER YA LAPTOP
Answered by
7
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)
Attachments:
Similar questions