Question

A ship leaves a port P and sails for 12km on a bearing 068. it then sails a further 20km on bearing of 106 to reach Q. what is the distance between P & Q. what is the bearing of Q from P.

Answer 1

Answer on question #72638 - Math - Trigonometry

A ship leaves a port P end sails for $12\mathrm{km}$ on a bearing 068. It then sails a further $20\mathrm{km}$ on bearing of 106 to reach Q. What is the distance between P and Q. What is the bearing of Q from P.

Solution

The angle $\mathrm{PSQ} = 180 - (106 - 68) = 144$ degree.

The angle $\mathrm{ASN} = 180 - 106 = 74$ degree

The angle $\mathrm{BSQ} =$ the angle $\mathrm{ASN} = 74$ degree

The angle $\mathrm{BQS} = 180 - 90 - 74 = 16$ degree

The angle $\mathrm{SPQ} = 180 - 144 - 16 = 20$ degree

The angle $\mathrm{QPN} = \mathrm{NPS} + \mathrm{CPQ} = 68 + 20 = 88$ degree

$\left| P Q \right| ^ {2} = \left| P S \right| ^ {2} + \left| S Q \right| ^ {2} - 2 * \left| P S \right| * \left| S Q \right| * \cos (P S Q) = 2 0 ^ {2} + 1 2 ^ {2} - 2 * 1 2 * 2 0 * \cos (1 4 4) = 4 0 0 + 1 4 4 - 4 8 0 * (- 0. 8 0 9) = 5 4 4 + 3 8 8 = 9 3 2. \left| P Q \right| = S Q R (9 3 2) = 3 0. 5 3 k m.$

Answer

$\left| P Q \right| = 30.53 \mathrm{~km}$ . The angle $\mathrm{QPN} = 88$ degree

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