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Q.

A harbour in the direction $60^{0}$ south-west from a fort and at a distance of 30 km from it. A ship sets out from the harbour at noon and sails due east at 10 km/hour, then the time (in hours) in the evening at which the ship will be 70 km from the fort is

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answer is 8.

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Detailed Solution

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Let A be the position of the harbour and O be of the fort. OA = 30, OB = 70. Let AB = x.
Question Image
Applying cosine formula in $Δ O A B$ , we get
$\cos 60^{0} = \frac{O A^{2} + A B^{2} - O B^{2}}{2 O A . A B} \Rightarrow \frac{1}{2} = \frac{900 + x^{2} - 4900}{2 \times 30 \times x}$
$\Rightarrow x^{2} - 30 x - 4000 = 0 \Rightarrow x = 80$
AB = 80 kms
Speed of ship = 10 km/hr
Time = 8hrs

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