Let's solve the triangle using the Law of Sines! The given triangle has:

• $a = 10$ (side opposite angle $A$)
• $b = 7$ (side opposite angle $B$)
• $\angle A = 42^\circ$

We need to find the following:

1. Angle $B$
2. Angle $C$
3. Side $c$

The Law of Sines states:

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

Step 1: Find $\angle B$ using the Law of Sines

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

Substitute the known values:

$\frac{10}{\sin 42^\circ} = \frac{7}{\sin B}$

First, calculate $\sin 42^\circ$:

$\sin 42^\circ \approx 0.6691$

Now, solve for $\sin B$:

$\frac{7}{10 / 0.6691} = \sin B$

$\sin B = \frac{7 \times 0.6691}{10} \approx 0.467$

Now, find $\angle B$ by taking the inverse sine:

$B = \sin^{-1}(0.467) \approx 27.9^\circ$

So, $\angle B \approx 27.9^\circ$.

Step 2: Find $\angle C$

Since the sum of the angles in a triangle is $180^\circ$:

$\angle C = 180^\circ - \angle A - \angle B$

$\angle C = 180^\circ - 42^\circ - 27.9^\circ \approx 110.1^\circ$

So, $\angle C \approx 110.1^\circ$.

Step 3: Find Side $c$ using the Law of Sines

We now use the Law of Sines again to find side $c$:

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

Substitute the known values:

$\frac{10}{\sin 42^\circ} = \frac{c}{\sin 110.1^\circ}$

First, calculate $\sin 110.1^\circ$:

$\sin 110.1^\circ \approx 0.936$

Now, solve for $c$:

$c = \frac{10 \times 0.936}{0.6691} \approx 14.0$

So, $c \approx 14.0$.

Final Solution

The solution for the triangle is:

• $\angle A = 42^\circ$
• $\angle B \approx 27.9^\circ$
• $\angle C \approx 110.1^\circ$
• $a = 10$
• $b = 7$
• $c \approx 14.0$

Do you need any more steps or further explanation?

Related Questions:

1. How can the Law of Sines be applied in non-obtuse triangles?
2. What happens if the calculated sine value gives no valid angle (ambiguous case)?
3. How would the Law of Cosines help in this triangle?
4. Can we solve the triangle if only one side and two angles are known?
5. Why is it important to use degrees consistently when using trigonometric functions?

Tip: Always double-check angles to ensure the sum of the triangle's angles is exactly $180^\circ$.