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Sladkaya [172]
3 years ago
13

The Questions are in the PDF! please help me

Mathematics
1 answer:
pshichka [43]3 years ago
6 0

Answer:

I'm sorry it tells me to download the PDF and I can't download PDF so the answer is B

Step-by-step explanation:

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Answer:

What's the question?

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Can someone help me this question?
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a port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling no
Rom4ik [11]

Answer:

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

\frac{sin A}{a} = \frac{sin B}{b}

Therefore,

\frac{sin 45}{6.25} = \frac{sin \alpha }{7.5}

α = Angle between radar and ship α

∴ α = 58.052

Where we put

\frac{sin 45}{6.25} = \frac{sin \alpha }{x} to get

\frac{x}{6.25} = \frac{sin \alpha }{sin 45} and differentiate to get

\frac{\frac{dx}{dt} }{6.25} = cos\alpha\frac{\frac{d\alpha }{dt}  }{sin 45}

\frac{15sin45 }{6.25cos\alpha } =\frac{d\alpha }{dt}  }= 3.208 degrees/second = 0.05599 rad/sec.

6 0
3 years ago
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