To solve for the unknown angles in triangle $ABC$, let's use the Law of Sines and analyze the possibility of multiple solutions.

Given Data:

• Angle $C = 61^\circ 40'$ = $61.67^\circ$ (in decimal degrees)
• Side $c = 33.4$
• Side $b = 22.9$

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Step 1: Using the Law of Sines

The Law of Sines states: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

From this, we can solve for $\sin B$: $\frac{\sin B}{b} = \frac{\sin C}{c}$ $\sin B = b \cdot \frac{\sin C}{c}$

Substitute the given values: $\sin B = 22.9 \cdot \frac{\sin(61.67^\circ)}{33.4}$

First, calculate $\sin(61.67^\circ)$: $\sin(61.67^\circ) \approx 0.8763$

Now substitute: $\sin B = 22.9 \cdot \frac{0.8763}{33.4} \approx 0.6006$

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Step 2: Find Angle $B$

Using the inverse sine function: $B = \arcsin(0.6006) \approx 36.88^\circ$

Rounding to the nearest whole number: $B \approx 37^\circ$

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Step 3: Determine Angle $A$

Since the sum of angles in a triangle is $180^\circ$: $A = 180^\circ - C - B$ $A = 180^\circ - 61.67^\circ - 36.88^\circ \approx 81.45^\circ$

Rounding to the nearest whole number: $A \approx 81^\circ$

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Step 4: Check for Possible Solutions

For a given $\sin B$, there could be two potential solutions for angle $B$:

1. $B_1 = \arcsin(0.6006) = 36.88^\circ$
2. $B_2 = 180^\circ - 36.88^\circ = 143.12^\circ$

However, if $B_2$ is used: $A = 180^\circ - C - B_2 = 180^\circ - 61.67^\circ - 143.12^\circ \approx -24.79^\circ$ Since a triangle cannot have a negative angle, there is only one possible solution.

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Final Results:

• One solution exists.
• $B \approx 37^\circ$
• $A \approx 81^\circ$

Would you like a step-by-step explanation of how the Law of Sines works?

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Additional Questions:

1. How do we calculate the area of triangle $ABC$ given its sides and angles?
2. How would the solution change if side $b$ were longer than side $c$?
3. What happens if the given angle $C$ is obtuse instead of acute?
4. Could we solve this problem using the Law of Cosines? Why or why not?
5. How can we verify the solution geometrically?

Tip: Always verify whether the second possible angle $B_2 = 180^\circ - B$ is valid to check for ambiguous cases in triangle problems.