To solve this problem, we will determine:

1. The missing side $x$ of the triangle using the Law of Cosines.
2. The missing angle $\theta$ using angle relationships in the triangle.

Step 1: Labeling the triangle

The triangle has:

• Side $AC = 10$,
• Side $AB = 8$,
• $\angle C = 40^\circ$,
• $x = BC$ (to find),
• $\theta = \angle A$ (to find).

Step 2: Solve for $x$ using the Law of Cosines

The Law of Cosines states: $c^2 = a^2 + b^2 - 2ab \cdot \cos(C),$ where $c$ is the side opposite angle $C$.

Substituting the values: $x^2 = 10^2 + 8^2 - 2(10)(8) \cdot \cos(40^\circ).$ Simplify: $x^2 = 100 + 64 - 160 \cdot \cos(40^\circ).$

Using $\cos(40^\circ) \approx 0.766$: $x^2 = 100 + 64 - 160 \cdot 0.766.$ $x^2 = 164 - 122.56.$ $x^2 = 41.44.$ $x \approx \sqrt{41.44} \approx 6.44.$

So, $x \approx 6.44$.

Step 3: Solve for $\theta$ (Angle A)

We now use the Law of Sines: $\frac{\sin(A)}{a} = \frac{\sin(C)}{c}.$

Substitute: $\frac{\sin(\theta)}{8} = \frac{\sin(40^\circ)}{6.44}.$

Using $\sin(40^\circ) \approx 0.643$: $\frac{\sin(\theta)}{8} = \frac{0.643}{6.44}.$ $\sin(\theta) = 8 \cdot \frac{0.643}{6.44}.$ $\sin(\theta) \approx 8 \cdot 0.0999.$ $\sin(\theta) \approx 0.799.$

Taking the inverse sine: $\theta \approx \arcsin(0.799) \approx 53.1^\circ.$

Final Answer:

• $x \approx 6.44$,
• $\theta \approx 53.1^\circ$.

Would you like more explanation on any step or assistance with related concepts?

Related questions to deepen your understanding:

1. How does the Law of Cosines differ from the Pythagorean theorem?
2. Why is the Law of Sines necessary to find the angle $\theta$?
3. What happens if the triangle is not acute—does the solution process change?
4. How would you find the third angle in the triangle using your results?
5. Can you solve this problem using only the Law of Sines? Why or why not?

Tip: Always double-check trigonometric calculations by verifying your results against the triangle's properties!