Question

You are on a fishing boat that leaves its pier and heads east. After traveling for 24 ​miles, there is a report warning of rough seas directly south. The captain turns the boat and follows a bearing of S40​°W for 13.5 miles. The pier is labeled point A, the turning point is labeled point B, and the ships current position is labeled C. Line segment A B lies on the east/west line where A is to the left of B and is labeled 24 miles. Line segment B C is 40 degrees west of south and is labeled 13.5 miles. Line segment C A is labeled b. There is a compass rose where north is up. 40 degrees 13.5 24 a. At this​ time, how far are you from the​ boat's pier? b. What bearing could the boat have originally taken to arrive at this​ spot?

Answer 1

a) b = 18.486 miles

b) Rac = S55°58'59.57"E

Explanation:

Given

AB = 24 miles

BC = 13.5 miles

Rbc = S40​°W

CA = b = ?

a) If α is the angle between AB and BC:

α = 90° - Rbc = 90° - 40° = 50°

⇒ α = 50°

We use The Law of Cosines as follows

CA² = b² = AB² + BC² - 2*AB*BC*Cos α

⇒ b² = (24 miles)² + (13.5 miles)² - 2*(24 miles)*(13.5 miles)*Cos 50°

⇒ b² = 341.72 miles²

⇒ b = √(341.72 miles²)

⇒ CA = b = 18.486 miles

b) We use The Law of Sines as follows

CA/Sin α = BC/Sin (90°- Rac)

⇒ Sin (90°- Rac) = BC*Sin α/CA

⇒ Sin (90°- Rac) = (13.5 miles)*Sin 50°/(18.486 miles)

⇒ Sin (90°- Rac) = 0.5594

⇒ 90°- Rac = Sin⁻¹(0.5594) = 34.017°

⇒ Rac = 90° - 34.017° = 55.983° = 55°58'59.57"

⇒ Rac = S55°58'59.57"E

Answer 2

(a)18.49 Miles

(b)
`S56^(0)E` or
`124^0`

Explanation:

(a)From the diagram, Angle B=50 degrees.

The distance of the boat at C from the boat's pier at A is the line AC.

We solve for AC using the Cosine rule.

`|AC|^2=|AB|^2+|BC|^2-2|AB||BC|Cos B\\=24^2+13.5^2-2(24)(13.5)Cos 50\\=341.72\\|AC|=√(341.72) \\|AC|=18.49 miles`

The distance of the boat from the pier is 18.49 miles.

(b)The bearing the boat could have originally taken to arrive at this​ spot is the bearing of C from A.

to calculate this, we find first the Angle at A.

Using Sine rule

`(a)/(Sin A) =(b)/(Sin B) \\(13.5)/(Sin A) =(18.49)/(Sin 50) \\18.49 X Sin A =13.5 X sin 50\\Sin A=13.5 X sin 50 / 18.49\\A = arcsin (13.5 X sin 50 / 18.49)\\A=34^0`

Therefore the bearing of C from A

`=90+34\\=124^0 \text{ (In 3-Digit Notation)}`

In Compass Bearing

Bearing of C from A

`=90^(0)-34^(0)\\= S56^(0)E`

The boat could have originally taken a bearing of
`S56^(0)E` degrees to arrive at C.