To solve this triangle with sides $a = 73$, $c = 67$, and angle $C = 40^\circ$, we can use the Law of Cosines and Law of Sines to find the missing angles $A$ and $B$, and side $b$.

Step 1: Solve for Side $b$ Using the Law of Cosines

The Law of Cosines formula is: $b^2 = a^2 + c^2 - 2ac \cdot \cos(C)$ Substitute the values: $b^2 = 73^2 + 67^2 - 2 \cdot 73 \cdot 67 \cdot \cos(40^\circ)$ Calculating $b$:

1. $73^2 = 5329$
2. $67^2 = 4489$
3. $2 \cdot 73 \cdot 67 = 9782$
4. $\cos(40^\circ) \approx 0.7660$

$b^2 = 5329 + 4489 - 9782 \cdot 0.7660$ $b^2 = 9818 - 7487.412 = 2330.588$ $b \approx \sqrt{2330.588} \approx 48.3$

So, $b \approx 48.3$.

Step 2: Solve for Angle $A$ Using the Law of Sines

The Law of Sines states: $\frac{\sin(A)}{a} = \frac{\sin(C)}{c}$ Rearrange to solve for $\sin(A)$: $\sin(A) = \frac{a \cdot \sin(C)}{c}$ Substitute the values: $\sin(A) = \frac{73 \cdot \sin(40^\circ)}{67}$ Calculating $A$:

1. $\sin(40^\circ) \approx 0.6428$
2. $73 \cdot 0.6428 = 46.9244$
3. $\sin(A) = \frac{46.9244}{67} \approx 0.7004$

$A \approx \arcsin(0.7004) \approx 44.4^\circ$

Step 3: Solve for Angle $B$ Using the Angle Sum of a Triangle

Since $A + B + C = 180^\circ$, $B = 180^\circ - A - C$ $B = 180^\circ - 44.4^\circ - 40^\circ \approx 95.6^\circ$

Summary of Results

• $A \approx 44.4^\circ$
• $B \approx 95.6^\circ$
• $b \approx 48.3$

Would you like more detailed steps or explanations?

Here are 5 related questions you might consider:

1. How does the Law of Cosines differ from the Law of Sines?
2. When is it appropriate to use the Law of Sines versus the Law of Cosines?
3. Can the Law of Sines produce ambiguous cases? If so, how do we resolve them?
4. What happens if the computed value of $\sin(A)$ is greater than 1?
5. How can the area of a triangle be calculated using trigonometric values?

Tip: Always check your computed angles to ensure they add up to 180°, as a quick verification step.