Question

Two ships leave a port at the same time .one goes 24 km per hour in the direction N 45° E and other travels 32km per hr in the direction S 75° E. Find the distance between the shops at the end of 3 hours .

Answer 1

Answer:

Answer is

$24\sqrt{2}$

Step-by-step explanation:

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Answer 2

Question :

Two ships leave a port at the same time .one goes 24 km per hour in the direction N 45° E and other travels 32km per hr in the direction S 75° E. Find the distance between the shops at the end of 3 hours .

${\red{\huge{\underline{\mathbb{Answer:-}}}}}$

To find : the distance between the shops at the end of 3 hours .

Given : one ship travels with speed 24 km/hr in direction N45°E

and the other ship travels with speed 32 km/hr in direction S 75° E.

${\purple{\huge{\underline{\mathbb{Theory }}}}}$

${\orange {\boxed{\large{\bold{The \: law \: of \: Cosines }}}}}$

Theorem :

In any ∆ABC,

$1)a {}^{2} = b {}^{2} + c {}^{2} - 2bc \cos(A)$

$2)b {}^{2} = c {}^{2} + a {}^{2} - 2ac \cos(B)$

$3)c {}^{2} = a {}^{2} + b {}^{2} - 2ab \cos(C)$

$\huge{\bold{ Solution : }}$

let P and Q be the postions of ships at the end of 3 hours . then ,

1st Ship goes 24Kmph for 3 hours

⇒OA = 3× 24 = 72 km

2nd Ship goes 32Kmph for 3 hours

⇒OQ = 3 × 32 = 96 Km

Now ,

by using Cosine formula in ∆OPQ ,

PQ² = OP² + OQ² -2 OP × OQ × cos 60°

⇒ PQ² = (72)² + (96)² - 2× 72× 96 × (1/2 )

⇒PQ² = 5184 + 9216 - 912

⇒PQ² = 7488

⇒ PQ = √ 7488 Km = 86.533 Km

Therefore , the distance between the ships at the end of 3 hours is 86.53 Km.