Let's solve part (a) using the sine law.

Given:

• Side $c = 25 \, \text{cm}$
• Angle $C = 35^\circ$
• Angle $B = 70^\circ$

We need to find side $b$.

Sine Law:

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

Since angle $A$ can be calculated as:

$A = 180^\circ - 70^\circ - 35^\circ = 75^\circ$

Now applying the sine law:

$\frac{b}{\sin(70^\circ)} = \frac{25}{\sin(35^\circ)}$

Solving for $b$:

$b = \frac{25 \cdot \sin(70^\circ)}{\sin(35^\circ)}$

Calculating the sines:

$\sin(70^\circ) \approx 0.940$ $\sin(35^\circ) \approx 0.574$

$b = \frac{25 \cdot 0.940}{0.574} \approx \frac{23.5}{0.574} \approx 40.94 \, \text{cm}$

Therefore, side $b \approx 40.94 \, \text{cm}$.

Would you like me to solve the other triangles as well or clarify any steps?

Related questions:

1. How do you derive the sine law from the area of a triangle?
2. How can the cosine law be used for non-right triangles?
3. Why is the sine law useful for oblique triangles?
4. What is the difference between the sine and cosine laws?
5. How do you verify the calculated side length using the triangle inequality?

Tip:

Always ensure the angles in a triangle add up to $180^\circ$ before applying the sine or cosine law.