A harbour H is a point of reference. Ship X is 16 km from H on a bearing of 090°. Ship Y is 14 km from H on a bearing of 130°. Ship Z is 26.31 km from H on a bearing of 240°. Find the distance between

The bearing of X from H is $090^\circ$.
The bearing of Y from H is $130^\circ$.
The angle between HX and HY is the difference in their bearings.

$$\angle XHY = 130^\circ - 090^\circ = 40^\circ$$

In $\triangle XHY$, we have $HX = 16$ km, $HY = 14$ km, and $\angle XHY = 40^\circ$.
The Cosine Rule states $c^2 = a^2 + b^2 - 2ab \cos C$.

$$XY^2 = HX^2 + HY^2 - 2(HX)(HY) \cos(\angle XHY)$$ $$XY^2 = 16^2 + 14^2 - 2(16)(14) \cos(40^\circ)$$ $$XY^2 = 256 + 196 - 448 \cos(40^\circ)$$ $$XY^2 = 452 - 448(0.7660)$$ $$XY^2 = 452 - 343.168$$ $$XY^2 = 108.832$$ $$XY = \sqrt{108.832}$$ $$XY \approx 10.432 \text{ km}$$

The distance between X and Y is approximately 10.43 km.

b) The distance from harbour

This part of the question is incomplete. It asks "The distance from harbour:" without specifying which ship. Based on the previous parts, it's likely asking for the distance of Z from the harbour, which was given in the problem statement.

The distance of Z from the harbour H is given as $HZ = 26.31$ km.
The distance from the harbour to Z is 26.31 km.

c) The distance between X and Z

The bearing of X from H is $090^\circ$.
The bearing of Z from H is $240^\circ$.
The angle between HX and HZ is the difference in their bearings.

$$\angle XHZ = 240^\circ - 090^\circ = 150^\circ$$

In $\triangle XHZ$, we have $HX = 16$ km, $HZ = 26.31$ km, and $\angle XHZ = 150^\circ$.

$$XZ^2 = HX^2 + HZ^2 - 2(HX)(HZ) \cos(\angle XHZ)$$ $$XZ^2 = 16^2 + (26.31)^2 - 2(16)(26.31) \cos(150^\circ)$$ $$XZ^2 = 256 + 692.2261 - 841.92 \cos(150^\circ)$$ $$XZ^2 = 948.2261 - 841.92(-0.8660)$$ $$XZ^2 = 948.2261 + 729.00992$$ $$XZ^2 = 1677.23602$$ $$XZ = \sqrt{1677.23602}$$ $$XZ \approx 40.954 \text{ km}$$

The distance between X and Z is approximately 40.95 km.

Final Answers:

a) The distance between X and Y is $\boxed{\text{10.43 km}}$

b) The distance from the harbour to Z is $\boxed{\text{26.31 km}}$

c) The distance between X and Z is $\boxed{\text{40.95 km}}$