#Help me with question 5 and 10 law of sines/cosines

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atomic vortex
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For 10 I got distance as 291.5

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atomic vortex
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A ship is cruising parallel to the coast of the speed of 18 mph spot of lighthouse is 15° off the starboard bow 15° to the right of straight ahead 15 minutes later the lighthouse is 22° off the starboard bow. How far is the ship from the shore?

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wet sonnet
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you got speed and you got time

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thus you can find the distance covered by the ship in 15 minutes

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let the lighthouse be AB and the ship be C

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so at first ACB is 15 degrees with the horizontal

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let x be the distance of the ship from the shore

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now using trigonometry you can find AB in terms of x

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that's our first equation

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and after 15 minutes, C makes an angle of 22 degrees

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here CB would be x minus the distance travelled by the ship in 15 minutes

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we got CB in terms of x, and using trigonometry we can find AB once again

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compare both the AB and you will get x