To solve the triangle using the Law of Sines, we use the formula:

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

Given:

• $A = 30.2^\circ$
• $a = 26$
• $b = 35$

We need to find:

• $B$
• $C$
• $c$

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Step 1: Solve for $\sin B$

Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$

Rearranging for $\sin B$: $\sin B = \frac{b \cdot \sin A}{a}$

Substitute the values: $\sin B = \frac{35 \cdot \sin(30.2^\circ)}{26}$

First, calculate $\sin(30.2^\circ)$: $\sin(30.2^\circ) \approx 0.502$

Now substitute: $\sin B = \frac{35 \cdot 0.502}{26} \approx \frac{17.57}{26} \approx 0.676$

Thus: $\sin B \approx 0.676$

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Step 2: Find $B$

To find $B$, take the arcsine of $\sin B$: $B = \arcsin(0.676)$

$B \approx 42.6^\circ$

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Step 3: Find $C$

The sum of angles in a triangle is $180^\circ$: $C = 180^\circ - A - B$

Substitute: $C = 180^\circ - 30.2^\circ - 42.6^\circ$

$C \approx 107.2^\circ$

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Step 4: Solve for $c$

Using the Law of Sines again: $\frac{a}{\sin A} = \frac{c}{\sin C}$

Rearranging for $c$: $c = \frac{a \cdot \sin C}{\sin A}$

Substitute the values: $c = \frac{26 \cdot \sin(107.2^\circ)}{\sin(30.2^\circ)}$

First, calculate $\sin(107.2^\circ)$: $\sin(107.2^\circ) \approx 0.973$

Substitute: $c = \frac{26 \cdot 0.973}{0.502} \approx \frac{25.298}{0.502} \approx 50.4$

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Final Results:

• $B \approx 42.6^\circ$
• $C \approx 107.2^\circ$
• $c \approx 50.4$

Would you like further clarification or a step-by-step walkthrough? Here are five related questions for further exploration:

1. How does the Law of Sines help solve oblique triangles?
2. What happens if $\sin B$ exceeds 1 during calculation?
3. Can we apply the Law of Sines in right triangles?
4. How do we determine if there are two possible solutions (ambiguous case)?
5. What is the relationship between the Law of Sines and the Law of Cosines?

Tip: Always double-check the sum of angles to confirm that it equals $180^\circ$.