Question

in right △abc with m∠c = 90°, m∠b = 75°, and ab = 12 cm. find the area of △abc. answer: cm² 1 attempt(s) used

Explanation:

Step1: Find angle A

In a triangle, the sum of 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 sides

In right - triangle $ABC$ with hypotenuse $AB = 12$ cm. $\sin B=\frac{AC}{AB}$ and $\cos B=\frac{BC}{AB}$.
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 $\cos75^{\circ}=\cos(45^{\circ}+30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$.
$AC = AB\sin B=12\sin75^{\circ}$ and $BC = AB\cos B = 12\cos75^{\circ}$.

Step3: Calculate the area of the triangle

The area of a right - triangle $S=\frac{1}{2}AC\times BC$.
Substitute $AC = 12\sin75^{\circ}$ and $BC = 12\cos75^{\circ}$ into the area formula:
$S=\frac{1}{2}\times12\sin75^{\circ}\times12\cos75^{\circ}$.
Using the double - angle formula $\sin2\alpha = 2\sin\alpha\cos\alpha$, then $\sin75^{\circ}\cos75^{\circ}=\frac{1}{2}\sin150^{\circ}$.
$S=\frac{1}{2}\times12\times12\times\frac{1}{2}\sin150^{\circ}$.
Since $\sin150^{\circ}=\frac{1}{2}$, then $S=\frac{1}{2}\times12\times12\times\frac{1}{2}\times\frac{1}{2}=18$ $cm^{2}$.

Answer:

$18$