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Ship X is 16 km from harbour 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°. Calculate the distance between X and Y.

Mathematics

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Answers

3 parts
a

10.43 km\text{10.43 km}

b

26.31 km\text{26.31 km}

c

40.95 km\text{40.95 km}

Step-by-step solution

4 steps

You are correct in identifying the ambiguity regarding ship Z's position.

Let H be the harbour.
Ship X is 16 km from H on a bearing of 090∘090^\circ. So, HX=16HX = 16 km.
Ship Y is 14 km from H on a bearing of 130∘130^\circ. So, HY=14HY = 14 km.
Ship Z is described with two pieces of information: "26.31 Km to the west of Y" and "on a bearing of 240∘240^\circ from the harbour". These two pieces of information are contradictory if interpreted strictly. For typical bearing problems, when a distance and bearing from the harbour are given for a ship, that distance refers to the distance from the harbour. Therefore, we will assume that the distance of Z from the harbour is HZ=26.31HZ = 26.31 km, and its bearing from the harbour is 240∘240^\circ. The information "26.31 Km to the west of Y" is considered extraneous or a misstatement, as it creates a contradiction with the given bearing from the harbour.

Part a· 2 steps

The distance between X and Y

  1. Find the angle ∠XHY\angle XHY.

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

    ∠XHY=130∘−090∘=40∘\angle XHY = 130^\circ - 090^\circ = 40^\circ

  2. Use the Cosine Rule to find the distance XY.

    In △XHY\triangle XHY, we have HX=16HX = 16 km, HY=14HY = 14 km, and ∠XHY=40∘\angle XHY = 40^\circ.
    The Cosine Rule states c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C.

    XY2=HX2+HY2−2(HX)(HY)cos⁡(∠XHY)XY^2 = HX^2 + HY^2 - 2(HX)(HY) \cos(\angle XHY) XY2=162+142−2(16)(14)cos⁡(40∘)XY^2 = 16^2 + 14^2 - 2(16)(14) \cos(40^\circ) XY2=256+196−448cos⁡(40∘)XY^2 = 256 + 196 - 448 \cos(40^\circ) XY2=452−448(0.7660)XY^2 = 452 - 448(0.7660) XY2=452−343.168XY^2 = 452 - 343.168 XY2=108.832XY^2 = 108.832 XY=108.832XY = \sqrt{108.832} XY≈10.432 kmXY \approx 10.432 \text{ km}

    The distance between X and Y is approximately 10.43 km\boxed{\text{10.43 km}}.

Part b

26.31 km\text{26.31 km}

The distance from the harbour

This part of the question asks for "The distance from harbour:". Based on the problem statement and our assumption for Z, this refers to the distance of Z from the harbour.
The distance of Z from the harbour H is given as HZ=26.31HZ = 26.31 km.
The distance from the harbour to Z is 26.31 km\boxed{\text{26.31 km}}.

Part c

40.95 km\text{40.95 km}

The distance between X and Z

  1. Find the angle ∠XHZ\angle XHZ.

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

    ∠XHZ=240∘−090∘=150∘\angle XHZ = 240^\circ - 090^\circ = 150^\circ

  2. Use the Cosine Rule to find the distance XZ.

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

    XZ2=HX2+HZ2−2(HX)(HZ)cos⁡(∠XHZ)XZ^2 = HX^2 + HZ^2 - 2(HX)(HZ) \cos(\angle XHZ) XZ2=162+(26.31)2−2(16)(26.31)cos⁡(150∘)XZ^2 = 16^2 + (26.31)^2 - 2(16)(26.31) \cos(150^\circ) XZ2=256+692.2261−841.92cos⁡(150∘)XZ^2 = 256 + 692.2261 - 841.92 \cos(150^\circ) XZ2=948.2261−841.92(−0.8660)XZ^2 = 948.2261 - 841.92(-0.8660) XZ2=948.2261+729.00992XZ^2 = 948.2261 + 729.00992 XZ2=1677.23602XZ^2 = 1677.23602 XZ=1677.23602XZ = \sqrt{1677.23602} XZ≈40.954 kmXZ \approx 40.954 \text{ km}

    The distance between X and Z is approximately 40.95 km\boxed{\text{40.95 km}}.

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