SOLUTION: A point Y is 400m north-west of X. A tree is north-east of Y and on a bearing 015° from X. Find the distance of the tree from
a) X
b) Y
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a) X
b) Y
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Question 1205425: A point Y is 400m north-west of X. A tree is north-east of Y and on a bearing 015° from X. Find the distance of the tree from
a) X
b) Y Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A point Y is 400m north-west of X. A tree is north-east of Y and on a bearing 015° from X. Find the distance of the tree from
a) X
b) Y
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Call the tree point T.
Assuming by NW you mean a bearing of 315 degs:
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Angle Y = 90 degs
Angle X = 60 degs
Angle T = 30 degs
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Use the Sine Law:
400/sin(30) = XT//sin(60)
800 = XT/sin(60) = XT*2/sqrt(3) = 800/sqrt(3) = ~461.88 meters
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XT = sqrt(XY^2 + YT^2) = ~ 516.4 meters
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Working with bearings (compass headings) can be confusing.
To change to angles wrt the x-axis, subtract from 450.
eg, 450 -315 --> 135 degs wrt the x-axis
450 - 15 ---> 435, so subtract 360 --> 75
450 - 135 = 315.
315 - 360 = -45
