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?
The yacht should take a bearing of the harbour, and then sail on that heading.
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°
Thanks, i got roughly the same answer, 282.
To find the bearing for the yacht to sail back to the harbor, we can use trigonometry and the concept of a bearing.
First, let's understand the given information. The yacht initially sails on a bearing of 065T for 3.4 km and then turns on a bearing of 127T for another 5 km.
To visualize this, imagine a coordinate plane. The initial bearing of 065T means the yacht is moving 65 degrees clockwise from true north. Starting from the harbor, the yacht sails for 3.4 km in this direction.
After that, the yacht turns on a bearing of 127T, which means it moves 127 degrees clockwise from true north. It sails for another 5 km in this direction.
Now, we need to find the bearing for the yacht to travel directly back to the harbor. To do this, we can use the concept of trigonometry.
Based on the information given, we have a right-angled triangle formed by the yacht's path back to the harbor, the distance covered in the first leg (3.4 km), and the distance covered in the second leg (5 km).
To find the bearing, we can calculate the angle formed by the path from the harbor to the yacht's current position. This can be done using inverse trigonometric functions.
Using the tangent ratio, we can find the angle:
tan(angle) = (opposite/adjacent)
tan(angle) = (3.4 km/5 km)
angle = arctan(3.4/5)
Taking the arctan of (3.4/5) gives us approximately 33.69 degrees. However, we need to note that the angle calculated is with respect to true north, and we need to convert it to a bearing.
To convert the angle to a bearing, we subtract it from 360 degrees because bearings are measured clockwise from true north. Thus, the bearing to sail directly back to the harbor is:
360 degrees - 33.69 degrees ≈ 326.31 degrees
Therefore, the yacht should sail on a bearing of approximately 326T (to the nearest minute) in order to travel directly back to the harbor.