A luxury liner leaves a port on a bearing of 110 degrees and travels 8.8 miles. It then turns due west and travels 2.4 miles. How far is the liner from the port, and what is its bearing from the port?

Distance= 6.6 miles

Bearing= N 62.854°W

Step-by-step explanation:

Let's determine angle b first

Angle b=20° (alternate angles)

Using cosine rule

Let the distance between the liner and the port be x

X² =8.8²+2.4²-2(8.8)(2.4)cos20

X²= 77.44 + 5.76-(39.69)

X²= 43.51

X= √43.51

X= 6.596

X= 6.6 miles

Let's determine the angles within the triangle using sine rule

2.4/sin b = 6.6/sin20

(2.4*sin20)/6.6= sin b

0.1244 = sin b

7.146= b°

Angle c= 180-20-7.146

Angle c= 152.854°

For the bearing

110+7.146= 117.146

180-117.146= 62.854°

Bearing= N 62.854°W

here is the

mark brainliest pl

correct answer:

b.(19,-2)