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Vlad [161]
3 years ago
12

A.)

Mathematics
1 answer:
photoshop1234 [79]3 years ago
8 0
I believe the answer is C.
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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
olga_2 [115]

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°

8 0
2 years ago
Read 2 more answers
Choose the most appropriate translation of the equation.Height(age) = 100A. The height of an oak tree (100) depends on the age o
Zigmanuir [339]
The height of an oak tree (which depends on the age of the tree) is 100
6 0
3 years ago
The graph of f(x) = is shown with g(x). Which equation represents the graph of g(x)? g(x) = g(x) = g(x) = + 1 g(x) = –1
tresset_1 [31]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Please hurry!!!!!!!!!!!!!!!!!!!!!!!!!!
DerKrebs [107]

Answer:

the radius of circle s is point,radius,arc,

3 0
3 years ago
Finding multiples that go into 86 52 18
Anton [14]
It's best to use the lowest number since the GCF will max out on the lowest number's highest factor.

18:
2
3
6
9
18

Let's start from the largest factor:

52/18 = 2.888888888_ (not a factor so 18 is eliminated)

52/9 = 5.7777777777_ (not a factor)

52/3 = 17.333333333_ (nope)

52/2 = 26

The only common factor within all of them is 2.
3 0
4 years ago
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