Question

A ship sails from harbour H on a bearing of
084° for 340 km until it reaches point P. It
then sails on a bearing of 210° for 160 km
until it reaches point Q.
a Calculate the distance between point Q
b On what bearing must the ship sail to
return directly to the harbour from Q?
and the harbour.

Answer 1

Given:

Distance from H to P = 340km

Distance from P to Q = 160km

Ship bearing angle H to P = 84°

Then sails from P to Q bearing angle = 210°

To find:

a) Calculate the distance between point Q  and the harbor.

b) On what bearing must the ship sail to  return directly to the harbor from Q?

Solution:

(b) Angle between  340km and 160km will be ∠HPQ

∠NPQ = 360° - 210° = 150° i.e total angle P                   - (1)

Now ∠NPH = 180° - 84° = 96°                                          - (2)

As we can see HN ║ PN and HP is transversal so sum of interior angle is 180°.

From (1) and (2) we get

∠HPQ = ∠NPQ - ∠NPH

∠HPQ = 150° - 96°

∠HPQ = 54°

(a) To calculate distance between Q and harbor 'H'

We will use cosine formula as we have one angle and two sides of ΔHPQ

∠P = 54° , HP = 340km , PQ = 160km

(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ

(QH)² = (340)² + (160)² - 2(340)(160)cos54°

(QH)² = 519904.965

QH = 227.8km ≈ 228km