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

Answer:

B.18. 44 miles

Step-by-step 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

$\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$

Using the sine law

$\frac{BC}{sin 40}=\frac{AB}{sin 100}$

$\frac{BC}{sin 40}=\frac{8}{sin 100}$

$BC=\frac{8\times sin40}{sin 100}$

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.