Question

find the angle b. do not use a calculator.
b = □°

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle ABC\), let \(a = 2\sqrt{6}\), \(b = 4\), and \(A = 60^{\circ}\). So, \(\frac{2\sqrt{6}}{\sin B}=\frac{4}{\sin 60^{\circ}}\).

Step2: Solve for \(\sin B\)

Cross - multiply: \(4\sin B=2\sqrt{6}\sin60^{\circ}\). Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), then \(4\sin B = 2\sqrt{6}\times\frac{\sqrt{3}}{2}\). Simplify the right - hand side: \(2\sqrt{6}\times\frac{\sqrt{3}}{2}=\sqrt{18}=3\sqrt{2}\). So, \(\sin B=\frac{3\sqrt{2}}{4}\) is incorrect. Wait, correct application: \(\frac{\sin B}{2\sqrt{6}}=\frac{\sin60^{\circ}}{4}\). Then \(\sin B=\frac{2\sqrt{6}\sin60^{\circ}}{4}\). Substitute \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\): \(\sin B=\frac{2\sqrt{6}\times\frac{\sqrt{3}}{2}}{4}=\frac{\sqrt{18}}{4}=\frac{3\sqrt{2}}{4}\) (wrong). Wait, correct formula: \(\frac{\sin B}{AC}=\frac{\sin A}{BC}\). \(AC = 2\sqrt{6}\), \(BC = 4\), \(A = 60^{\circ}\). \(\sin B=\frac{AC\sin A}{BC}\). \(\sin B=\frac{2\sqrt{6}\times\frac{\sqrt{3}}{2}}{4}=\frac{\sqrt{18}}{4}=\frac{3\sqrt{2}}{4}\) (error). Wait, correct: \(\frac{\sin B}{4}=\frac{\sin60^{\circ}}{2\sqrt{6}}\). \(\sin B=\frac{4\sin60^{\circ}}{2\sqrt{6}}=\frac{4\times\frac{\sqrt{3}}{2}}{2\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{6}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\).

Step3: Determine the angle \(B\)

Since \(\sin B=\frac{\sqrt{2}}{2}\) and \(A = 60^{\circ}\), and in a triangle, \(B\) is acute (because \(AC>BC\) in the side - angle relationship: larger side is opposite larger angle. Here, if \(B = 135^{\circ}\), \(A + B>180^{\circ}\)). So \(B = 45^{\circ}\).

Answer:

\(45\)