Two ships leave a port at the same time. One | vedclass

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(C) Let the port be at the origin $O(0,0)$.After $2 	ext{ hours}$,the first ship $A$ is at a distance of $8 	imes 2 = 16 	ext{ km}$ from the port i

Vedclass · Accepted Answer

$8 \sqrt{19}$

Vedclass · Answer

(C) Let the port be at the origin $O(0,0)$.After $2 	ext{ hours}$,the first ship $A$ is at a distance of $8 	imes 2 = 16 	ext{ km}$ from the port in the direction $E 50^{\circ} N$.The second ship $B$ is at a distance of $12 	imes 2 = 24 	ext{ km}$ from the port in the direction $S 20^{\circ} E$.The angle between the two directions is calculated as follows:The direction $E 50^{\circ} N$ is $50^{\circ}$ north of the east axis.The direction $S 20^{\circ} E$ is $20^{\circ}$ east of the south axis.The angle between the east axis and the south axis is $90^{\circ}$.Thus,the total angle $	heta$ between the two ships is $50^{\circ} + (90^{\circ} - 20^{\circ}) = 50^{\circ} + 70^{\circ} = 120^{\circ}$.Using the Law of Cosines in $	riangle OAB$ where $OA = 16$,$OB = 24$,and $\angle AOB = 120^{\circ}$:$AB^2 = OA^2 + OB^2 - 2(OA)(OB) \cos(120^{\circ})$$AB^2 = 16^2 + 24^2 - 2(16)(24)(-0.5)$$AB^2 = 256 + 576 + 384 = 1216$$AB = \sqrt{1216} = \sqrt{64 	imes 19} = 8 \sqrt{19} 	ext{ km}$.

Two ships leave a port at the same time. One of them moves in the direction of $E 50^{\circ} N$ with a speed of $8 \text{ kmph}$ and the other moves in the direction of $S 20^{\circ} E$ with a speed of $12 \text{ kmph}$. Then the distance between the ships at the end of $2 \text{ hours}$ is (in km).

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