Question

find the length of side a.
c
120°
3
c
a
15°
b
b = 45°
a = ?
c =

Explanation:

Step1: Find angle at B

In triangle \(ABC\), sum of angles is \(180^\circ\). So \(\angle B = 180^\circ - 120^\circ - 15^\circ = 45^\circ\) (matches given \(B = 45^\circ\)).

Step2: Apply Law of Sines

Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, side \(b = 3\) (opposite \(\angle B = 45^\circ\)), \(\angle A = 15^\circ\), \(\angle B = 45^\circ\), side \(a\) opposite \(\angle A\). So \(\frac{a}{\sin 15^\circ}=\frac{3}{\sin 45^\circ}\).

Step3: Solve for \(a\)

\(a=\frac{3\sin 15^\circ}{\sin 45^\circ}\). Calculate \(\sin 15^\circ=\sin(45^\circ - 30^\circ)=\sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588\), \(\sin 45^\circ=\frac{\sqrt{2}}{2}\approx0.7071\). Then \(a=\frac{3\times0.2588}{0.7071}\approx1.1\) (or exact: \(\frac{3(\sqrt{6}-\sqrt{2})}{4\sqrt{2}}=\frac{3(\sqrt{3}-1)}{4}\approx1.1\)).

Answer:

\(\approx 1.1\) (or exact form \(\frac{3(\sqrt{3}-1)}{4}\))