Question

A ship is stationary at sea. A tugboat is 28 km away at a bearing of 315°, and a yacht is 21 km from the tugboat at a bearing of 210°. Draw a scale diagram showing the positions of the three vessels. Use a scale of 1 : 350,000. Measure the distance from the ship to the yacht to the nearest km.​

Answer 1

Answer:

ans is 33.6

Step-by-step explanation:

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Answer 2

The distance of the yacht from the ship is found by accurately drawing

the relative location of the three vessels.

Response:

• The measured distance of the yacht from the ship is approximately 30 km.

Which drawing method can be used in which the distance of the yacht from the ship can be measured?

The given parameters are;

Distance of the tugboat from the ship, a = 28 km

Bearing of tugboat from the ship = 315°

Distance of the yacht from the tugboat, b = 21 km

Bearing of the yacht from the = 210°

Using a scale of 1 : 350,000, we have;

$The \ drawing \ of \ a = \mathbf{\dfrac{28,000}{350,000} }= 0.08$

The length of a in the drawing = 0.08 m = 8 cm

$The \ drawing \ of \ b = \mathbf{\dfrac{21,000}{350,000}} = 0.06$

Which gives;

b in the drawing = 0.06 m = 6 cm

Using the above dimensions and directions, the drawing of the relative

location of the three vessels can be accurately created using MS Word.

From the application, the vector form of the distance, d, of the ship from the yacht is presented as follows;

• d = 8.655·i + 0.445·j

Which gives;

d = √(8.655² + 0.445²) = 8.67

Which gives;

• Distance of the yacht from the ship on the drawing is d ≈ 8.67 cm = 0.0867 m

Actual distance = 0.0867 m × 350,000 = 30,345 m = 30.345 km ≈ 30 km

• The distance of the yacht from the ship ≈ 30 km

Using cosine rule, where the angle formed at the tugboat = 75°, we have;

d² = 28² + 21² - 2 × 28 × 21 × cos(75°) ≈ 920.63

Which gives;

d ≈ √(920.63) ≈ 30

The distance of the yacht from the ship, d ≈ 30 km

Learn more about bearings in mathematics here:

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