Question

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now? 3)How far is the ship from its starting point? 4)On what bearing is it now from its oringinal position?​

Answer 1

Explanation:

1.

let the distance in north direction=x

total distance=250+x km

`(x)/(150) =cos 75\\x=150 cos 75\approx38.82 km\\`

so distance north=240+38.82=278.82 km

2.let the distance to north=y

`(y)/(150) =sin 75\\y=150 sin 75 \approx 144.89`

so distance in East direction=144.89 km.

3.

`(38.82)/(278.82) =tan \alpha \\\alpha =tan^(-1)((38.82)/(278.82) )\approx7.93^\circ`

Answer 2

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

`\sf \sin(\theta)=(O)/(H)\quad\cos(\theta)=(A)/(H)\quad\tan(\theta)=(O)/(A)`

where:

• 
  `\theta` is the angle
• O is the side opposite the angle
• A is the side adjacent the angle
• H is the hypotenuse (the side opposite the right angle)

Cosine rule

`c^2=a^2+b^2-2ab \cos C`

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

`\implies \sf \cos(75^(\circ))=(y)/(150)`

`\implies \sf y=150\cos(75^(\circ))`

`\implies \sf y=38.92285677`

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

`\implies \sf \sin(75^(\circ))=(x)/(150)`

`\implies \sf x=150\sin(75^(\circ))`

`\implies \sf x=144.8888739`

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

`\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)`

`\implies \sf d=√(250^2+150^2-2(250)(150) \cos (180-75))`

`\implies \sf d=323.1275729`

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

`\implies \sf Bearing=180^(\circ)+\tan^(-1)\left((Total\:Eastern\:distance)/(Total\:Northern\:distance)\right)`

`\implies \sf Bearing=180^(\circ)+\tan^(-1)\left((150\sin(75^(\circ)))/(250+150\cos(75^(\circ)))\right)`

`\implies \sf Bearing=180^(\circ)+26.64077...^(\circ)`

`\implies \sf Bearing=207^(\circ)`

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°