Question

A ship is stationary at sea.

Answer 1

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 x.

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 a = 6 and c = 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.