Help me solve (5+6i)^2
2 answers:
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The bearing of R from P is 019 and the distance |PR| is 8.75
<h3>The complete question</h3>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.
<h3>The diagram that represents the information</h3>See attachment
<h3>The bearing of R from P</h3>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
<h3>The distance |PR|</h3>In (b), we have:
PR = 8.75
Hence, the distance |PR| is 8.75
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