Asked by Trig
A yacht sets sail from a harbour and sails on a bearing of 065T for 3.4km. it then turns and sails on a bearing of 127T for another 5km. On what bearing (to the nearest minute) should the yacht sail if it was to travel directly back to the harbour?
Answers
Answered by
Steve
The yacht should take a <u>bearing</u> of the harbour, and then sail on that <u>heading</u>.
Not sure what 065T means, but I assume it's 65° clockwise from True North. If so, then the heading back to the harbour is found via
Label the diagram as follows:
H = harbour
P = point of turning
Q = final location
If the sides opposite these angles of triangle HPQ are h,p,q respectively, then
p^2 = 3.4^2 + 5^2 - 2*3.4*5*cos(138°) = 61.83
p = 7.86
Then to find angle Q,
sinQ/3.4 = sin138°/7.86
Q = 16.82° ≈ 17°
So the bearing of the harbour from Q is 360-(63+17) = 280°
Not sure what 065T means, but I assume it's 65° clockwise from True North. If so, then the heading back to the harbour is found via
Label the diagram as follows:
H = harbour
P = point of turning
Q = final location
If the sides opposite these angles of triangle HPQ are h,p,q respectively, then
p^2 = 3.4^2 + 5^2 - 2*3.4*5*cos(138°) = 61.83
p = 7.86
Then to find angle Q,
sinQ/3.4 = sin138°/7.86
Q = 16.82° ≈ 17°
So the bearing of the harbour from Q is 360-(63+17) = 280°
Answered by
Trig
Thanks, i got roughly the same answer, 282.
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