Question

1. two ships leave an island at the same time. the first travels at a bearing of n42°e for 530 miles. the second travels at a bearing of s78°w for 490 miles. how far apart are the ships? (round to the nearest mile.)

Explanation:

Step1: Find the angle between paths

First ship: N42°E (42° east of north). Second ship: S78°W (78° west of south). The angle between them is 42° + 78° + 90° = 210°? No, correct angle: from north, first is 42° right, second is 180°+78°=258° from north. Difference: 258°-42°=216°? No, smaller angle: 360°-216°=144°. So angle θ=144°.

Step2: Apply Law of Cosines

Let a=530, b=490, θ=144°. Law of Cosines: $c^2=a^2+b^2-2ab\cosθ$.
Calculate: $a^2=530^2=280900$, $b^2=490^2=240100$, $2ab=2×530×490=519400$, $\cos144°≈-0.8090$.
$c^2=280900+240100-519400×(-0.8090)=521000 + 519400×0.8090≈521000 + 420204.6≈941204.6$.
$c≈\sqrt{941204.6}≈970$.

Answer:

970