The problem provided involves solving a triangle using the given side lengths and angle:

• $b = 65$
• $c = 39$
• $B = 59^\circ$

We need to solve for the remaining angles $A$ and $C$, and the side $a$, using the law of sines.

Step 1: Apply the Law of Sines

$\frac{b}{\sin B} = \frac{c}{\sin C} = \frac{a}{\sin A}$

Step 2: Solve for $C$

From the law of sines, we can solve for $C$ using the relationship between $b$, $c$, and their corresponding angles: $\frac{\sin C}{c} = \frac{\sin B}{b}$ Substitute the known values: $\frac{\sin C}{39} = \frac{\sin 59^\circ}{65}$

Calculate $\sin 59^\circ$ (using four decimal places): $\sin 59^\circ \approx 0.8572$ Now solve for $\sin C$: $\sin C = \frac{39 \times 0.8572}{65} = 0.5145$

Thus, $C = \sin^{-1}(0.5145) \approx 30.98^\circ$ Rounding to the nearest tenth: $C \approx 31.0^\circ$

Step 3: Solve for $A$

Since the sum of angles in a triangle is $180^\circ$: $A = 180^\circ - B - C = 180^\circ - 59^\circ - 31.0^\circ = 90.0^\circ$

Step 4: Solve for $a$

Now, use the law of sines again to find $a$: $\frac{a}{\sin A} = \frac{b}{\sin B}$ Since $\sin A = \sin 90^\circ = 1$, we get: $a = \frac{65}{\sin 59^\circ} \approx \frac{65}{0.8572} \approx 75.8$

Final Answer:

• $A = 90.0^\circ$
• $C = 31.0^\circ$
• $a \approx 75.8$

Let me know if you need further clarification or details on any step.

Further Related Questions:

1. How does the law of sines work in non-right triangles?
2. What happens if there is no solution to the triangle based on the given information?
3. Can you use the law of cosines to solve this triangle? How would that change the steps?
4. Why do we carry intermediate computations to four decimal places?
5. What is the importance of rounding angles to the nearest tenth?

Tip:

Always check the triangle's angle sum to ensure the computed angles make sense, especially when solving ambiguous cases.