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Two ships leave  a port from a point at the same time. one goes with a velocity of 3km/h along North-east  making an angle of $45^{0}$ with east direction and the other travels with a velocity of 4km/h along South-East making an angle of $15^{0}$ with East direction . Then, the distance  between  the ships  at the end of two hours is 

Answer:

Explanation:

Distance travelled by ship OA direction in 2h is 6km and OB  direction in 2h 138 km

 $AB^{2}= OA^{2}+OB^{2}-2OAOB \cos \angle AOB$

 $AB^{2}=(6)^{2}+(8)^{2}-2 \times 6 \times 8 \cos 60^{0}$

 $AB^{2}=36+64-2 \times 6 \times 8 \times \frac{1}{2}$

 $AB^{2}=100-48$

 $AB= \sqrt{100-48}=\sqrt{52}$

AB= $2\sqrt{13}$