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Nookie1986 [14]
2 years ago
5

Please Help Me!!! √36/169 = ?

Mathematics
1 answer:
LuckyWell [14K]2 years ago
6 0

Answer:

Step-by-step explanation:

0.03550295857

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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
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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 <u>30 km</u>.

<h3>Which drawing method can be used in which the distance of the yacht from the ship can be measured?</h3>

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 <em>a</em> in the drawing = 0.08 m = 8 cm

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

Which gives;

<em>b</em> in the drawing = 0.06 m = 6 cm

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

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From the application, the vector form of the distance, <em>d</em>, 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 ≈ <u>30 km</u>

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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