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bekas [8.4K]
3 years ago
6

A ship is stationary at sea.

Mathematics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

The distance between the yacht from the ship is about 33.6 km.

Step-by-step explanation:

The diagram is attached below.

Point A is the shipe, Point B is the tugboat, and Point C is the yacht.

Since the tugboat is 28 km away, and the scale being used is 1 cm : 4 km, the tugboat is 7 cm away from the ship at a 210° bearing on the diagram.

Likewise, the yacht is 6 cm away from the tugboat at a 210° bearing.

To find the distance between the yacht to the ship, we need to determine the value of <em>x</em>.

To do so, we can use the Law of Cosines:

b^2=a^2+c^2-2ac\cos(B)

First, we need to determine ∠B.

A has a bearing of 210°. 180° is covered by QI and IV. So, the small angle in QIII is 30°.

By Alternate Interior Angles, the angle in QI (inside the triangle) of Point B must also be 30°.

And since Point B has a bearing of 310°, 270° is covered in total by QI, QIV, and QIII. So, 40° is left in QII.

Therefore, 50° is the measure of the angle in QIII within the triangle of Point B.

Thus, ∠B measures 50° + 30° = 80°.

So, ∠B measures 80°. We also know that <em>a</em> = 6 and <em>c</em> = 7. Substitute:

b^2=(6)^2+(7)^2-2(6)(7)\cos(80)

Simplify and take the square root of both sides. So:

b=\sqrt{85-84\cos(80)}

Approximate using a calculator:

b=8.3912...\approx 8.39\text{ cm}

Since 1 cm : 4km, to find the distance in km, we simply need to multiply by four. So:

\Rightarrow b=4(8.3912...)=33.5651\approx 33.6\text{ km}

The distance between the yacht from the ship is about 33.6 km.

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The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

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=

b

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For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

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 sin  B =

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The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

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h

b

or

h = c  sin  B       a n d         h = b  sin  C

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c  sin  B = b  sin  C

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c

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(see Supplementary angles trig identities)

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a

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b

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