Question

solve the right triangle.
write your answers in simplified, rationalized form. do not round.
bc=
ac=
m∠c=

Explanation:

Step1: Find \(m\angle C\)

In a right - triangle, the sum of the two non - right angles is \(90^{\circ}\).
Since \(\angle A = 90^{\circ}\) and \(\angle B=60^{\circ}\), then \(m\angle C=90^{\circ}-\angle B\).
\(m\angle C = 90^{\circ}-60^{\circ}=30^{\circ}\)

Step2: Find \(BC\) (hypotenuse)

We know that \(\cos B=\frac{AB}{BC}\). Given \(AB = 5\sqrt{3}\) and \(\angle B = 60^{\circ}\), and \(\cos60^{\circ}=\frac{1}{2}\).
From \(\cos B=\frac{AB}{BC}\), we can rewrite it as \(BC=\frac{AB}{\cos B}\).
Substitute \(AB = 5\sqrt{3}\) and \(\cos B=\cos60^{\circ}=\frac{1}{2}\) into the formula: \(BC=\frac{5\sqrt{3}}{\frac{1}{2}}=10\sqrt{3}\)

Step3: Find \(AC\)

We know that \(\tan B=\frac{AC}{AB}\). Given \(AB = 5\sqrt{3}\) and \(\angle B = 60^{\circ}\), and \(\tan60^{\circ}=\sqrt{3}\).
From \(\tan B=\frac{AC}{AB}\), we can rewrite it as \(AC = AB\times\tan B\).
Substitute \(AB = 5\sqrt{3}\) and \(\tan B=\sqrt{3}\) into the formula: \(AC=5\sqrt{3}\times\sqrt{3}=15\)

Answer:

\(BC = 10\sqrt{3}\), \(AC = 15\), \(m\angle C=30^{\circ}\)