Question

in right $\triangle abc$ with $mangle c = 90^{circ}, mangle b = 75^{circ}$, and $ab = 12$ cm. find the area of $\triangle abc$. answer: $square$ $cm^{2}$

Explanation:

Step1: Find angle A

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

Step2: Use trigonometric relations

$\sin B=\frac{AC}{AB}$, $\cos B=\frac{BC}{AB}$. Given $AB = 12$ cm, $AC = AB\sin B=12\sin75^{\circ}$, $BC = AB\cos B=12\cos75^{\circ}$.

Step3: Calculate the area of the right - triangle

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

Answer:

$18$