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Question

Difficulty: MediumSine and Cosine Rules

Two fishing boats leave a harbor HH at the same time. Boat AA travels on a bearing of 020∘020^\circ for a distance of 8 km8\text{ km}, while Boat BB travels on a bearing of 140∘140^\circ for a distance of 7 km7\text{ km}. What is the distance between Boat AA and Boat BB in kilometers?

Answer: 13 km

Answer

13 km
The distance between the two boats forms the third side of triangle HABHAB, where HA=8 kmHA = 8\text{ km}, HB=7 kmHB = 7\text{ km}, and the included angle at the harbor HH is ∠AHB=140∘−020∘=120∘\angle AHB = 140^\circ - 020^\circ = 120^\circ. By the Cosine Rule, AB2=82+72−2(8)(7)cos⁡(120∘)=64+49−112(−0.5)=169AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169. Taking the square root gives AB=13 kmAB = 13\text{ km}.

Step-by-Step Solution

1
Calculate the included angle between the direction vectors of the two boats from the harbor.
Included angle ∠AHB=140∘−020∘=120∘\angle AHB = 140^\circ - 020^\circ = 120^\circ
The angle between two bearings originating from the same point is the difference between their bearing angles.
2
State the Cosine Rule for side ABAB in triangle HABHAB.
AB2=HA2+HB2−2⋅HA⋅HB⋅cos⁡(∠AHB)AB^2 = HA^2 + HB^2 - 2 \cdot HA \cdot HB \cdot \cos(\angle AHB)
The Cosine Rule calculates an unknown side when two sides and their included angle (SAS) are given.
3
Substitute given side lengths HA=8 kmHA = 8\text{ km}, HB=7 kmHB = 7\text{ km}, and angle ∠AHB=120∘\angle AHB = 120^\circ.
AB2=82+72−2(8)(7)cos⁡(120∘)=64+49−112(−0.5)=169AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169
Since 120∘120^\circ is in the second quadrant, cos⁡(120∘)=−0.5\cos(120^\circ) = -0.5, which changes the subtracted term to addition.
4
Compute the principal square root of 169.
AB=169=13 kmAB = \sqrt{169} = 13\text{ km}
Distance is a non-negative scalar quantity.

Key Concept

Applying the Cosine Rule to solve bearing problems
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Topics

Sine and Cosine Rules

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