A ship sails from harbour H on a bearing of084° for 340 km until it r

1.	A ship sails from harbour H on a bearing of084° for 340 km until it reaches point P. Itthen sails on a bearing of 210° for 160 kmuntil it reaches point Q.a Calculate the distance between point Qb On what bearing must the ship sail toreturn directly to the harbour from Q?and the harbour.
Answer» 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

A ship sails from harbour H on a bearing of084° for 340 km until it reaches point P. Itthen sails on a bearing of 210° for 160 kmuntil it reaches point Q.a Calculate the distance between point Qb On what bearing must the ship sail toreturn directly to the harbour from Q?and the harbour.

Answer»

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