A ship leaves the port of Miami with a bearing of S80°E and a speed of 15 knots. After 1 hour, the ship turns 90° toward the south. After 2 hours, maintain the same speed. What is the bearing to the ship from port?

The bearing is N 55.62° W

Step-by-step explanation:

ship leaves the port of Miami with a bearing of S80°E and a speed of 15 knots.

It then turns 90° towards the south after one hour.

Still maintain the same speed and direction for two hours.

The bearing is just the angle difference from the ship current location to where it started.

Let the speed be km/h

Distance covered in the first round

= 15*1

= 15km

Distance covered in the second round

=15*2

= 30 km

Angle at C = (90-80)+90

Angle at C = 10+90= 100

Let the distance between the port and the ship be c

C²= a² + b² -2abcos

C²= 15²+30²-2(15)(30)cos 100

C²= 225+900+156.28

C²= 1281.28

C= 35.8 km

Using sine formula

30/sin x= 35.8/sin 100

30/35.8 * sin 100 = sinx

0.838*0.9848= sin x

0.8253= sin x

Sin ^-1 0.8253 = x

55.62° = x

The bearing is N 55.62° W