Question

17.a solve the following problems. given: △abc, m∠a = 60°, m∠c = 45°, ab = 9 find: the perimeter and area of △abc answer: clear all p_abc = a_abc =

Explanation:

Step1: Find angle B

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

Step2: Use the sine - rule to find side lengths

By the sine - rule \(\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{AC}{\sin B}\).
Since \(AB = 9\), \(\sin A=\sin60^{\circ}=\frac{\sqrt{3}}{2}\), \(\sin C=\sin45^{\circ}=\frac{\sqrt{2}}{2}\), and \(\sin B=\sin75^{\circ}=\sin(45^{\circ} + 30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\).
From \(\frac{AB}{\sin C}=\frac{BC}{\sin A}\), we have \(BC=\frac{AB\sin A}{\sin C}=\frac{9\times\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}=\frac{9\sqrt{3}}{\sqrt{2}}=\frac{9\sqrt{6}}{2}\).
From \(\frac{AB}{\sin C}=\frac{AC}{\sin B}\), we have \(AC=\frac{AB\sin B}{\sin C}=\frac{9\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}}=\frac{9(\sqrt{6}+\sqrt{2})}{2\sqrt{2}}=\frac{9(\sqrt{3} + 1)}{2}\).

Step3: Calculate the perimeter \(P_{\triangle ABC}\)

\(P_{\triangle ABC}=AB + BC+AC=9+\frac{9\sqrt{6}}{2}+\frac{9(\sqrt{3}+1)}{2}=\frac{18 + 9\sqrt{6}+9\sqrt{3}+9}{2}=\frac{27 + 9\sqrt{6}+9\sqrt{3}}{2}\).

Step4: Calculate the area \(A_{\triangle ABC}\)

The area formula is \(A=\frac{1}{2}AB\cdot AC\cdot\sin A\).
Substitute \(AB = 9\), \(AC=\frac{9(\sqrt{3}+1)}{2}\), and \(\sin A=\frac{\sqrt{3}}{2}\) into the formula.
\(A_{\triangle ABC}=\frac{1}{2}\times9\times\frac{9(\sqrt{3}+1)}{2}\times\frac{\sqrt{3}}{2}=\frac{81(3 + \sqrt{3})}{8}\).

Answer:

\(P_{\triangle ABC}=\frac{27 + 9\sqrt{6}+9\sqrt{3}}{2}\), \(A_{\triangle ABC}=\frac{81(3+\sqrt{3})}{8}\)