Question

A boat travels 8 miles north from point A to point B. Then it moves in the direction S 40°W and reaches point Finally, it turns S 40°E and returns to point A

The total distance covered by the boat is______miles

A. 14.95
B. 18.44
C. 20.04
D. 25.88

Answer 1

B.18. 44 miles

Explanation:

We are given that

Distance between A and B=8 miles

Angle B=Angle BCQ=40 degree (Alternate interior angles)

Angle ACB=180-Angle ACP-Angle BCQ

Angle ACB=180-40-40=100 degree

In triangle ABC

Angle A+ Angle B +Angle C=180 degree using sum of angles of triangle property

Substitute the values

`\angle A+40+100=180`

`\angle A+140=180`

`\angle A=180-140`

`\angle A=40` degree

Angle A=Angle B

When two angles are equal of a triangle then the triangle is isosceles triangle.

Therefore, triangle ABC is an isosceles triangle.

`\implies BC=AC`

Now, Sine law

`(a)/(sin A)=(b)/(sin B)=(c)/(sin C)`

Using the sine law

`(BC)/(sin 40)=(AB)/(sin 100)`

`(BC)/(sin 40)=(8)/(sin 100)`

`BC=(8* sin40)/(sin 100)`

BC=5.22

AC=BC=5.22 miles

Now, total distance covered by the boat=AB+BC+AC

Total distance covered by the boat=8+5.22+5.22=18.44 miles

Hence, option B is correct.