Question

In triangle ABC, angle A is 30 degrees, angle B is 45 degrees, and side a (opposite angle A) is 10 units long. What is the length of side b (opposite angle B)?

Answer 1

To find the length of side \( b \) in triangle \( ABC \), we can use the Law of Sines, which states:

\[
\frac{a}{\sin A} = \frac{b}{\sin B}
\]

Given:

- \( a = 10 \) (side opposite angle \( A \))
- \( \angle A = 30^\circ \)
- \( \angle B = 45^\circ \)

We need to find \( \sin A \) and \( \sin B \):

\[
\sin A = \sin 30^\circ = \frac{1}{2}
\]
\[
\sin B = \sin 45^\circ = \frac{\sqrt{2}}{2}
\]

Now we can set up the equation using the Law of Sines:

\[
\frac{10}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{2}}{2}}
\]

Calculating the left side:

\[
\frac{10}{\frac{1}{2}} = 10 \times 2 = 20
\]

So we have:

\[
20 = \frac{b}{\frac{\sqrt{2}}{2}}
\]

To solve for \( b \), multiply both sides by \( \frac{\sqrt{2}}{2} \):

\[
b = 20 \times \frac{\sqrt{2}}{2}
\]

Calculating \( b \):

\[
b = 10\sqrt{2}
\]

Thus, the length of side \( b \) is

\[
\boxed{10\sqrt{2}}
\]