Question

A cyclist set out from a town P on a bearing of to a town Q, 5 km away. He then moves on a bearing of to a town R, 6km from Q.

Answer 1

The bearing of R from P is 019 and the distance |PR| is 8.75

The complete question

A cyclist set out from a town P on a bearing of 060 to a town Q, 5 km away. He then moves on a bearing of 345 to a town R, 6km from Q.

The diagram that represents the information

See attachment

The bearing of R from P

Start by calculating the distance PR using the following cosine ratio

PR² = PQ² + QR² - 2 * PQ * PR cos(Q)

Substitute known values

PR² = 5² + 6² - 2 * 5 * 6 cos(105)

Evaluate

PR² = 76.53

Take the square root of both sides

PR = 8.75

Next, calculate angle R using the following sine ratio

PQ/sin(R) = PR/sin(Q)

Substitute known values

5/sin(R) = 8.75/sin(105)

Evaluate the quotient

5/sin(R) = 9.06

Multiply both sides by sin(R)

9.06 * sin(R) = 5

Divide both sides by 9.06

sin(R) = 0.5519

Take the arc sin of both sides

R = sin⁻¹(0.5519)

Evaluate

R = 34

Calculate angle P using:

P = 180 - 34 - 105

P = 41

The bearing of R from P is then calculated as:

Bearing = 060 - 41

Evaluate

Bearing =  019

Hence, the bearing of R from P is 019

The distance |PR|​

In (b), we have:

PR = 8.75

Hence, the distance |PR| is 8.75

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