A surveyor walking on a level ground at 10 Kph in the eastward direction sighted the top of a tall narra tree in the direction N12degreeE. After 30 mins the same tree now bears N48degreesE. How far was the surveyor from the tree during the first sight?
2007-10-13 22:19:58 · 3 answers · asked by delixir_21 1 in Science & Mathematics ➔ Mathematics
My teach said that he was going east
2007-10-14 01:39:50 · update #1
Yes he is going east.
We have 2 right triangles with one side is distance from road to the tree that is equal to H. And two other sides L1 and L2 such that L1 - L2 = 5km
H/L1 = tan(12) so L1 = H/tan(12)
H/L2 = tan(48) so L2 = H/tan(48)
H/tan(12) - H/tan(48) = 5
L = 5/(1/tan(12) - 1/tan(48))
We are looking for hypothenuse of the first triangle with side H and angle 12, so L = H/sin(12) and finally
H = 5/(sin(12)(1/tan(12) - 1/tan(48)))
2007-10-21 20:54:11 · answer #1 · answered by Alexey V 5 · 0⤊ 0⤋
The surveyor would have to be walking west. In half an hour
at 10 kph he would walk 5 km.
Let
A = surveyor's first sight location
B = surveyor's second sight location
T = location of tree
m
Now use the Law of Sines.
AT/sin42° = 5/sin36°
AT = 5(sin42°/sin36°) â 5.692 km
__________
The gentleman above had the right idea, but worked out the distance from the surveyor's second sight location.
2007-10-13 22:56:22 · answer #2 · answered by Northstar 7 · 0⤊ 1⤋
direction of walking might be going west . . please check it
the other angle = 180 - 42 - 102 = 36 deg
by sine law
dist = 5 sin 102 / sin 36 = 8.321 Km.
2007-10-13 22:32:16 · answer #3 · answered by CPUcate 6 · 0⤊ 1⤋