Questions: Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that angle B1 is larger than angle B2.) a=32, c=42, angle A = 33° angle B1 = square°, angle B2 = square° angle C1 = C2 = C2 = b° b1 = b2

Solution Steps

To solve the triangle using the Law of Sines, follow these steps:

1. Use the Law of Sines: Start by using the Law of Sines to find angle $C$. The formula is $\frac{a}{\sin A} = \frac{c}{\sin C}$.

2. Calculate Angle $C$: Rearrange the formula to solve for $\sin C$ and then find $C$ using the inverse sine function.

3. Determine Possible Triangles: Since the sine function can have two possible angles (an acute and an obtuse angle), check for both possibilities of angle $C$.

4. Find Angle $B$: Use the fact that the sum of angles in a triangle is $180^\circ$ to find angle $B$ for each possible triangle.

5. Calculate Side $b$: Use the Law of Sines again to find side $b$ for each triangle.

Using the Law of Sines, we have:

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

Substituting the known values:

$$\frac{32}{\sin(33^\circ)} = \frac{42}{\sin C}$$

Calculating $\sin C$:

$$\sin C = \frac{42 \cdot \sin(33^\circ)}{32} \approx 0.7148$$

Since $\sin C \approx 0.7148$, we can find the angles $C_1$ and $C_2$:

$$C_1 = \arcsin(0.7148) \approx 45.4^\circ$$ $$C_2 = 180^\circ - C_1 \approx 134.6^\circ$$

Using the triangle angle sum property $A + B + C = 180^\circ$:

For $C_1$:

$$B_1 = 180^\circ - 33^\circ - 45.4^\circ \approx 101.6^\circ$$

For $C_2$:

$$B_2 = 180^\circ - 33^\circ - 134.6^\circ \approx 12.4^\circ$$

Using the Law of Sines again to find side $b$:

For $B_1$:

$$b_1 = \frac{a \cdot \sin(B_1)}{\sin(A)} = \frac{32 \cdot \sin(101.6^\circ)}{\sin(33^\circ)} \approx 56.5$$

For $B_2$:

$$b_2 = \frac{a \cdot \sin(B_2)}{\sin(A)} = \frac{32 \cdot \sin(12.4^\circ)}{\sin(33^\circ)} \approx 14.5$$

Final Answer