Question

in right △abc with m∠c = 90°, m∠b = 75°, and ab = 12 cm. find the area of △abc. answer: □ cm²

Explanation:

Step1: Find angle A

In a right - triangle, the sum of the interior angles is 180°. Given $\angle C = 90^{\circ}$ and $\angle B=75^{\circ}$, then $\angle A=180^{\circ}-\angle C - \angle B=180^{\circ}-90^{\circ}-75^{\circ}=15^{\circ}$.

Step2: Use trigonometric relations to find the legs

Let $AB = c = 12$ cm. We know that $\sin B=\frac{AC}{AB}$ and $\sin A=\frac{BC}{AB}$. So $AC = c\sin B=12\sin75^{\circ}$ and $BC = c\sin A = 12\sin15^{\circ}$.
Since $\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Step3: Calculate the area of the triangle

The area of a right - triangle $S=\frac{1}{2}AC\times BC$. Substituting $AC = 12\sin75^{\circ}$ and $BC = 12\sin15^{\circ}$, we get $S=\frac{1}{2}\times12\sin75^{\circ}\times12\sin15^{\circ}$.
Since $\sin75^{\circ}\sin15^{\circ}=\frac{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}{16}=\frac{6 - 2}{16}=\frac{1}{4}$, then $S=\frac{1}{2}\times12\times12\times\frac{1}{4}=18$ $cm^{2}$.

Answer:

$18$