Two ships leave a port at the same time. One goes $24 k m$ per hour in the direction ${N 45}^{0} E$ and other travels $32 k m$ per hour in the direction ${S 75}^{0} E$ . Find the distance between the ships at the end of $3$ hours.

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Solution

$1 s t S h i p g o e s 24 K m p h f o r 3 h o u r s,$

$O A = \frac{24 k m}{h r} * 3 h r = 72 k m$

$S o, T h e 1 s t s h i p t r a v e l s 72 k m$

$2 n d S h i p g o e s 32 K m p h f o r 3 h o u r s,$

$O B = \frac{32 k m}{h r} * 3 h r = 96 k m$

$S o, T h e 2 n d S h i p t r a v e l s 96 K m$

$∴ ∠ A O B = 45 ° + (90 ° + 75 °) = 60 °$

$S o, A B^{2} = O A^{2} + O B^{2} - 2 \times O A \times O B \times C o s 60 °$

$= 72^{2} + 96^{2} - (2 \times 72 \times 96 \times \frac{1}{2})$

$A B = \sqrt{72^{2} + 96^{2} - 72 \times 96} = 86.53 K m$

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Two ships leave a port at the same time. One goes 24 km per hour in the direction N 450 E and other travels 32 km per hour in the direction S 750 E. Find the distance between the ships at the end of 3 hours.

1stShipgoes24Kmphfor3hours,OA=24kmhr∗3hr=72kmSo,The1stshiptravels72km2ndShipgoes32Kmphfor3hours,OB=32kmhr∗3hr=96kmSo,The2ndShiptravels96Km∴∠AOB=45°+(90°+75°)=60°So,AB2=OA2+OB2−2×OA×OB×Cos60°=722+962−(2×72×96×12)AB=√722+962−72×96=86.53Km

Two ships leave a port at the same time. One goes $24 k m$ per hour in the direction ${N 45}^{0} E$ and other travels $32 k m$ per hour in the direction ${S 75}^{0} E$ . Find the distance between the ships at the end of $3$ hours.