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algol [13]
2 years ago
9

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?​
Mathematics
2 answers:
olga_2 [115]2 years ago
8 0

Answer:

1)  288.8 km due North

2)  144.9 km due East

3)  323.1 km

4)  207°

Step-by-step explanation:

<u>Bearing</u>: The angle (in degrees) measured clockwise from north.

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{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)

<u>Cosine rule</u>

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.

<u>Question 1</u>

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})=\dfrac{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.

<u>Question 2</u>

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})=\dfrac{x}{150}

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

\implies \sf x=144.8888739

Therefore, the ship is now 144.9 km due East.

<u>Question 3</u>

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=\sqrt{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.

<u>Question 4</u>

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(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)

\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{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°

Arturiano [62]2 years ago
5 0

Answer:

Step-by-step explanation:

1.

let the distance in north direction=x

total distance=250+x km

\frac{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

\frac{y}{150} =sin 75\\y=150 sin 75 \approx 144.89

so distance in East direction=144.89 km.

3.

\frac{38.82}{278.82} =tan \alpha \\\alpha =tan^{-1}(\frac{38.82}{278.82} )\approx7.93^\circ

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