To solve this problem, we can apply the Law of Sines, given that we have two sides $a = 21.47$ cm, $b = 10.2$ cm, and one angle $A = 37.2^\circ$. Here’s the process step-by-step:

Step 1: Calculate angle $B$ using the Law of Sines

The Law of Sines states: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$ Rearranging to solve for $B$: $\sin(B) = \frac{b \cdot \sin(A)}{a}$

Step 2: Substitute the values and calculate $B$

$\sin(B) = \frac{10.2 \cdot \sin(37.2^\circ)}{21.47}$

1. Calculate $\sin(37.2^\circ) \approx 0.6046$.
2. Substitute and compute: $\sin(B) = \frac{10.2 \cdot 0.6046}{21.47} \approx 0.2871$

Step 3: Find angle $B$

Now, use the inverse sine function: $B = \sin^{-1}(0.2871) \approx 16.7^\circ$

Since $\sin(B)$ has two possible solutions in the range of $0^\circ$ to $180^\circ$ (one acute and one obtuse), $B$ could also be: $B' = 180^\circ - 16.7^\circ \approx 163.3^\circ$

Step 4: Check if each solution for $B$ is possible

1. If $B = 16.7^\circ$: Then angle $C$ can be found using: $C = 180^\circ - A - B = 180^\circ - 37.2^\circ - 16.7^\circ \approx 126.1^\circ$ This is a valid solution.

2. If $B = 163.3^\circ$: Then angle $C$ would be: $C = 180^\circ - A - B = 180^\circ - 37.2^\circ - 163.3^\circ \approx -20.5^\circ$ Since a triangle cannot have a negative angle, this solution is invalid.

Step 5: Find side $c$ for the valid solution

With $B \approx 16.7^\circ$ and $C \approx 126.1^\circ$, we can use the Law of Sines to find $c$: $\frac{c}{\sin(C)} = \frac{a}{\sin(A)}$ Rearrange and substitute values: $c = \frac{a \cdot \sin(C)}{\sin(A)} = \frac{21.47 \cdot \sin(126.1^\circ)}{\sin(37.2^\circ)}$

1. Calculate $\sin(126.1^\circ) \approx 0.8290$.
2. Substitute and compute: $c = \frac{21.47 \cdot 0.8290}{0.6046} \approx 29.43 \text{ cm}$

Conclusion

Only one solution is possible:

• Angle $B \approx 16.7^\circ$
• Angle $C \approx 126.1^\circ$
• Side $c \approx 29.43$ cm

The correct answer is therefore Option B.

Would you like further details or have any questions?

Related Questions

1. How does the Law of Sines determine whether a triangle has one or two solutions?
2. What happens if $\sin(B)$ exceeds 1 when solving for $B$?
3. Why does having an angle $C$ with a negative value mean a triangle is impossible?
4. How can we verify our solution once all angles and sides are known?
5. What scenarios require using the Law of Cosines instead of the Law of Sines?

Tip:

When working with ambiguous cases (SSA triangles), always check for two potential angle solutions for $B$ to determine if multiple triangles can be formed.