Let's calculate the area of the triangle using the given information.

Given:

• $AB = 10$ km (from point $A$ to $B$)
• Bearings:
  • $\angle CAB = 100^\circ$
  • $\angle ABC = 160^\circ$
  • $\angle BCA = 100^\circ$

Since all internal angles in the triangle sum to $360^\circ - 160^\circ - 100^\circ = 100^\circ$.

Step 1: Use the Law of Sines

We know that:

$\frac{AB}{\sin(\angle BCA)} = \frac{BC}{\sin(\angle CAB)} = \frac{CA}{\sin(\angle ABC)}$

We know $AB = 10$ km. Let's denote $BC = c$ and $AC = b$. We can set up the ratios:

$\frac{10}{\sin(100^\circ)} = \frac{c}{\sin(100^\circ)}$

But since we need the area, let’s directly calculate using:

Step 2: Area of Triangle Using Two Sides and Included Angle

$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle CAB)$

Given that $AC = AB = 10$, we get:

$\text{Area} = \frac{1}{2} \times 10 \times 10 \times \sin(100^\circ)$

Now calculate the sine and solve:

$\text{Area} = 0.5 \times 100 \times \sin(100^\circ)$

$\sin(100^\circ) \approx 0.9848$

$\text{Area} = 50 \times 0.9848 = 49.24 \, \text{km}^2$

So, the area of the triangle $ABC$ is approximately 49.24 square kilometers.

Would you like to discuss any other details or clarifications?

Further Questions

1. What are the exact lengths of sides $BC$ and $AC$?
2. Can you find the perimeter of the triangle?
3. How does the bearing affect the internal angles of a triangle?
4. What is the shortest distance between any two points?
5. How would the area change if any side was altered?

Tip

Always verify angles and calculations when working with bearings and trigonometry to avoid mistakes in complex problems.