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Let's analyze the problem:

### Given Information:
1. Town F is **50 km east** of Town G.
2. Town H is on a **bearing of 040° from Town F**.
3. The distance from Town F to Town H is **65 km**.

We need to:
1. Calculate the distance \( GH \) (the straight-line distance from Town G to Town H).
2. Calculate the bearing of Town H from Town G.

---

### Step 1: Represent the Problem as a Triangle
The triangle \( GFH \):
- \( GF = 50 \, \text{km} \) (eastward).
- \( FH = 65 \, \text{km} \) (at a bearing of 040° from F, so the angle between \( GF \) and \( FH \) is \( 90^\circ - 40^\circ = 50^\circ \)).

### Step 2: Use the Cosine Rule to Find \( GH \)
The formula for the cosine rule is:
\[
c^2 = a^2 + b^2 - 2ab \cos(C)
\]
Here:
- \( a = GF = 50 \),
- \( b = FH = 65 \),
- \( C = 50^\circ \) (angle between them).

Substitute into the formula:
\[
GH^2 = 50^2 + 65^2 - 2(50)(65)\cos(50^\circ)
\]

Calculate:
\[
GH^2 = 2500 + 4225 - 2(50)(65)(0.6428)
\]
\[
GH^2 = 2500 + 4225 - 4189.2
\]
\[
GH^2 = 2535.8
\]

\[
GH = \sqrt{2535.8} \approx 50.4 \, \text{km}
\]

So, the distance \( GH \) is approximately **50 km** to the nearest kilometer.

---

### Step 3: Find the Bearing of H from G
We use the Sine Rule to find the angle \( \theta \) at \( G \):
\[
\frac{\sin(\theta)}{FH} = \frac{\sin(C)}{GH}
\]

Substitute:
\[
\frac{\sin(\theta)}{65} = \frac{\sin(50^\circ)}{50.4}
\]
\[
\sin(\theta) = 65 \times \frac{0.7660}{50.4} \approx 0.9881
\]

\[
\theta \approx \arcsin(0.9881) \approx 81^\circ
\]

Now, the bearing of H from G:
- Start from north (0°) and rotate clockwise.
- Add \( \theta = 81^\circ \) to 90° (since \( GH \) is northeast).

\[
\text{Bearing} = 90^\circ + 81^\circ = 171^\circ
\]

---

### Final Answers:
1. The actual distance \( GH \) is approximately **50 km**.
2. The bearing of H from G is approximately **171°**.

Let me know if you want further clarification or additional details.

---

### Related Questions:
1. What is the relationship between bearings and angles in navigation problems?
2. How do you determine angles from bearings in trigonometry-based problems?
3. Can you apply the sine rule to directly calculate the bearing of one point from another?
4. What would happen if the bearing from F were changed (e.g., 045° instead of 040°)?
5. How can the cosine rule be used to check calculations in similar problems?

**Tip:** Always sketch the triangle to visualize the angles and sides when solving bearing problems.

Town F is 50 km east of town G. Town H is on a bearing of 040° from town F. The distance from F to H is 65 km. Calculate, to the nearest kilometer, the actual distance GH. Calculate, to the nearest degree, the bearing of H from G.

Learn how to solve a bearings problem involving towns F, G, and H. This solution uses trigonometry to calculate the distance GH and the bearing of H from G, applying the cosine and sine rules. Suitable for Grades 10-12.

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