Asked by Kalia
From the top of a 50m high bridge, two boats are seen at anchor. One boat is S 50degrees W and has an angle of depression of 38 degrees. The other boat is S 60degrees E and has a 35 degree angle of depression. How far apart are the boats?
Answers
Answered by
Henry
We draw 2 rt triangles with a common ver side:
1. Draw a rectangle with the long sides hor.
2. Draw a diagonal from the upper rt
corner to the lower lt corner. This is
the hyp of the larger triangle.The angle bet. the hyp and hor is 35 deg.
3. Draw a 2nd hyp from the upper rt corner to a 2nd point on the hor side.
This is the hyp of the smaller triangle. The dist. bet. the 2 hyp(d1)
is the dist. bet. the 2 boats. The dist
bet. the 2nd hyp and the ver side is d2. The angle bet. the 2nd hyp and hor
is 38 deg.
d1 + d2 = d = total dist. from the
bridge.
tan38 = 50/d2,
d2 = 50/tan38 = 64m.
tan35 = 50/(d1+d2),
d1+d2 = 50/tan35 = 71.4m.
d1 + d2 = 71.4,
d1 = 71.4 - d2 = 71.4 - 64 = 7.4m = dist. bet. boats.
1. Draw a rectangle with the long sides hor.
2. Draw a diagonal from the upper rt
corner to the lower lt corner. This is
the hyp of the larger triangle.The angle bet. the hyp and hor is 35 deg.
3. Draw a 2nd hyp from the upper rt corner to a 2nd point on the hor side.
This is the hyp of the smaller triangle. The dist. bet. the 2 hyp(d1)
is the dist. bet. the 2 boats. The dist
bet. the 2nd hyp and the ver side is d2. The angle bet. the 2nd hyp and hor
is 38 deg.
d1 + d2 = d = total dist. from the
bridge.
tan38 = 50/d2,
d2 = 50/tan38 = 64m.
tan35 = 50/(d1+d2),
d1+d2 = 50/tan35 = 71.4m.
d1 + d2 = 71.4,
d1 = 71.4 - d2 = 71.4 - 64 = 7.4m = dist. bet. boats.
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