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3 years ago

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4. A ship traveled a distance of 50 km from port O at a bearing of 120°, then another 100 km to port B at a bearing of 200 °. Wh

at is the distance between ports O and B?

1 answer:

Ivanshal [37]3 years ago

8 0

Step-by-step explanation:

Jesu

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Question in picture<br><br>A) 16/63<br><br>B)-16/63<br><br>C) 63/16<br><br>D) -63/16

Orlov [11]

ANSWER
$\tan(x + y) = - \frac{63}{16}$

EXPLANATION

We were given that,

$\csc(x) = \frac{5}{3}$

This implies that,

$\sin(x) = \frac{3}{5}$

We use the Pythagorean identity

$\sin^{2} (x) + \cos^{2} (x)= 1$
to get,

$\cos(x) = \sqrt{1 - ( { \frac{3}{5} })^{2}} = \frac{4}{5}$

We were also given that,

$\cos(y) = \frac{5}{13}$

This means that,

$\sin(y) = \sqrt{1 - {( \frac{5}{13}) }^{2} } = \frac{12}{13}$

This is because,

$0 < \: x \: < \frac{\pi}{2}$

$0 < \: y \: < \frac{\pi}{2}$

This angles are in the first quadrant so we pick the positive values.

$\tan(x + y) = \frac{ \sin(x + y) }{ \cos(x + y) }$

$\tan(x + y) = \frac{ \sin(x ) \cos(y) + \sin(y) \cos(x) }{ \cos(x) \cos(y) - \sin(x) \sin(y) }$

$\tan(x + y) = \frac{ \frac{3}{5} \times \frac{5}{13} + \frac{12}{13} \times \frac{4}{5} }{ \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} }$

$\tan(x + y) = - \frac{63}{16}$

The correct answer is D

4 0

3 years ago