Answer:
(a)18.49 Miles
(b)[tex]S56^{0}E[/tex] or [tex]124^0[/tex]
Step-by-step 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.
[tex]|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|=\sqrt{341.72} \\|AC|=18.49 miles[/tex]
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
[tex]\frac{a}{Sin A} =\frac{b}{Sin B} \\\frac{13.5}{Sin A} =\frac{18.49}{Sin 50} \\18.49 X Sin A =13.5 X sin 50\\Sin A=13.5 X sin 50 \div 18.49\\A = arcsin (13.5 X sin 50 \div 18.49)\\A=34^0[/tex]
Therefore the bearing of C from A
[tex]=90+34\\=124^0 \text{ (In 3-Digit Notation)}[/tex]
In Compass Bearing
Bearing of C from A
[tex]=90^{0}-34^{0}\\= S56^{0}E[/tex]
The boat could have originally taken a bearing of [tex]S56^{0}E[/tex] degrees to arrive at C.
Answer:
a) b = 18.486 miles
b) Rac = S55°58'59.57"E
Step-by-step 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
