4. A cruise ship leaves dock T and sails south 125 km. Then it travels 235 km on a bearing of 230° to island V. What is the distance from the island to the dock? N T 125 km 230° 235 km

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A: In △ACD, tan 12∘=DCACAC=450tan 12∘=4500.2125≈2117 ft

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A: Topic:- triangle and trigonometry

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Q: 9. 5. 6. 49° 11 C = ? 52 C 40° 14 A A

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Q: 4

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A: Please refer to the image below.

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A: We have to calculate final distance of the ship from the port.

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A: Given: