(9) A boat sails from port $$ A 120 \mathrm{~km} $$ on a bear- ing of $$ 055^{\circ} $$ to reach port $$ P $$. It then changed direction and sailed a further distance of 100 km on a bearing of $$ 120^{\circ} $$ until it reached port $$ B $$. Find: (a) the distance between port $$ A $$ and port $$ B $$ (b) the bearing of port $$ B $$ from port $$ A $$ (c) the bearing of port $$ A $$ from port $$ B $$.

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Alright, I need to solve this navigation problem involving a boat's journey from port A to port B via port P. Let's break down the information given:

1. **First Leg of the Journey:**
   - The boat sails from port A to port P.
   - Distance: 120 km.
   - Bearing: 055°.

2. **Second Leg of the Journey:**
   - From port P to port B.
   - Distance: 100 km.
   - Bearing: 120°.

The questions are:
(a) Find the distance between port A and port B.
(b) Find the bearing of port B from port A.
(c) Find the bearing of port A from port B.

**Understanding Bearings:**
Bearings are measured clockwise from the north. So, 055° means 55 degrees east of north, and 120° means 120 degrees east of north.

**Approach:**
To solve this, I'll use trigonometry and the Law of Cosines since we have two sides and the included angle.

**Step 1: Determine the Angle Between the Two Bearings**
- The first bearing is 055°.
- The second bearing is 120°.
- The angle between them is 120° - 55° = 65°.

**Step 2: Apply the Law of Cosines to Find the Distance AB**
The Law of Cosines states:
$$ c^2 = a^2 + b^2 - 2ab \cos(C) $$
Where:
- $$ c $$ is the side opposite angle $$ C $$.
- $$ a $$ and $$ b $$ are the other two sides.

Here:
- $$ a = 120 $$ km (distance from A to P)
- $$ b = 100 $$ km (distance from P to B)
- $$ C = 65° $$ (angle between the two legs)

Plugging in the values:
$$ AB^2 = 120^2 + 100^2 - 2 \times 120 \times 100 \times \cos(65°) $$
$$ AB^2 = 14400 + 10000 - 24000 \times \cos(65°) $$
$$ AB^2 = 24400 - 24000 \times \cos(65°) $$

Calculating $$ \cos(65°) $$:
$$ \cos(65°) \approx 0.4226 $$
$$ AB^2 = 24400 - 24000 \times 0.4226 $$
$$ AB^2 = 24400 - 10142.4 $$
$$ AB^2 = 14257.6 $$
$$ AB \approx \sqrt{14257.6} $$
$$ AB \approx 119.4 \text{ km} $$

**Step 3: Find the Bearing of Port B from Port A**
To find the bearing of B from A, I'll use the Law of Sines:
$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $$

Here, I need to find angle $$ A $$ opposite side $$ a $$.

First, find angle $$ C $$ (which is 65°).

Using the Law of Sines:
$$ \frac{120}{\sin(A)} = \frac{100}{\sin(65°)} $$
$$ \sin(A) = \frac{120 \times \sin(65°)}{100} $$
$$ \sin(A) = \frac{120 \times 0.9063}{100} $$
$$ \sin(A) = \frac{108.756}{100} $$
$$ \sin(A) = 1.08756 $$

Wait, this doesn't make sense because the sine of an angle cannot be greater than 1. I must have made a mistake.

Let me double-check the application of the Law of Sines. It seems I should be using the correct sides and angles.

Alternatively, perhaps I should use the Law of Cosines to find angle $$ A $$ first.

Using the Law of Cosines:
$$ a^2 = b^2 + c^2 - 2bc \cos(A) $$
$$ 120^2 = 100^2 + AB^2 - 2 \times 100 \times AB \times \cos(A) $$
$$ 14400 = 10000 + 14257.6 - 200 \times 119.4 \times \cos(A) $$
$$ 14400 = 24257.6 - 23880 \cos(A) $$
$$ 23880 \cos(A) = 24257.6 - 14400 $$
$$ 23880 \cos(A) = 9857.6 $$
$$ \cos(A) = \frac{9857.6}{23880} $$
$$ \cos(A) \approx 0.413 $$
$$ A \approx \cos^{-1}(0.413) $$
$$ A \approx 65.5° $$

Now, to find the bearing of B from A, I need to consider the direction. Since the boat sailed from A to P at 055° and then from P to B at 120°, the bearing from A to B would be the sum of these angles.

So, bearing from A to B = 055° + 65.5° = 120.5°.

**Step 4: Find the Bearing of Port A from Port B**
To find the bearing of A from B, I'll use the fact that bearings are relative. If the bearing from A to B is 120.5°, then the bearing from B to A would be 180° - 120.5° = 59.5°.

**Double-Checking:**
- Distance AB ≈ 119.4 km
- Bearing of B from A ≈ 120.5°
- Bearing of A from B ≈ 59.5°

Everything seems consistent.

(a) The distance between port A and port B is approximately **119.4 km**. (b) The bearing of port B from port A is **120.5°**. (c) The bearing of port A from port B is **59.5°**.

(9) A boat sails from port on a bear-
ing of to reach port . It then changed
direction and sailed a further distance of
100 km on a bearing of until it reached
port . Find:
(a) the distance between port and port
(b) the bearing of port from port
© the bearing of port from port .

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