Question

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

By the sine - rule, $\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{AC}{\sin B}$.
We know $AB = 9$, $\sin A=\sin60^{\circ}=\frac{\sqrt{3}}{2}$, $\sin C=\sin45^{\circ}=\frac{\sqrt{2}}{2}$, 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 $AC=\frac{AB\sin B}{\sin C}=\frac{9\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}}=\frac{9(\sqrt{6}+\sqrt{2})}{4}\times\frac{2}{\sqrt{2}}=\frac{9(\sqrt{3} + 1)}{2}$.

Step3: Calculate the perimeter

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

Step4: Calculate the area

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

Answer:

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