Question

12.a solve the following problems. given: △abc, m∠a = 60°, m∠c = 45°, ab = 12 find: the perimeter of △abc answer: p_abc =

Explanation:

Step1: Find angle B

The sum of angles in a triangle is 180°. So, $\angle B=180^{\circ}-\angle A - \angle C=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.

Step2: Use the sine - rule to find side BC

By the sine - rule $\frac{BC}{\sin A}=\frac{AB}{\sin C}$. Substituting the given values: $\frac{BC}{\sin60^{\circ}}=\frac{12}{\sin45^{\circ}}$. Then $BC = \frac{12\sin60^{\circ}}{\sin45^{\circ}}=\frac{12\times\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 6\sqrt{6}$.

Step3: Use the sine - rule to find side AC

By the sine - rule $\frac{AC}{\sin B}=\frac{AB}{\sin C}$. First, $\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$. Then $\frac{AC}{\sin75^{\circ}}=\frac{12}{\sin45^{\circ}}$, so $AC=\frac{12\sin75^{\circ}}{\sin45^{\circ}}=\frac{12\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}}= 3(\sqrt{6}+\sqrt{2})$.

Step4: Calculate the perimeter

The perimeter $P_{ABC}=AB + BC+AC$. Substitute the values: $P_{ABC}=12 + 6\sqrt{6}+3(\sqrt{6}+\sqrt{2})=12 + 6\sqrt{6}+3\sqrt{6}+3\sqrt{2}=12 + 9\sqrt{6}+3\sqrt{2}$.

Answer:

$12 + 9\sqrt{6}+3\sqrt{2}$