sosossossossos

a) Ta có $\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$ (tổng 3 góc trong 1 tam giác) $\Rightarrow \widehat{C}=180^{\circ}-\widehat{A}-\widehat{B}=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$ Ta có $\frac{AB}{AC}=\cos A$ $\Rightarrow AB=AC\times \cos A=10\times \cos 30^{\circ}=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}$ Ta có $\frac{BC}{AC}=\sin A$ $\Rightarrow BC=AC\times \sin A=10\times \sin 30^{\circ}=10\times \frac{1}{2}=5$ b) Diện tích tam giác ABC là: $S_{ABC}=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 5\sqrt{3}\times 5=\frac{25\sqrt{3}}{2}$ Đáp số: a) $\widehat{C}=60^{\circ}; AB=5\sqrt{3}; BC=5$ b) $S_{ABC}=\frac{25\sqrt{3}}{2}$ a) Ta có $\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$ (tổng 3 góc trong 1 tam giác) $\Rightarrow \widehat{B}=180^{\circ}-\widehat{A}-\widehat{C}=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$ Ta có $\frac{AC}{AB}=\sin B$ $\Rightarrow AC=AB\times \sin B=10\times \sin 30^{\circ}=10\times \frac{1}{2}=5$ Ta có $\frac{BC}{AB}=\cos B$ $\Rightarrow BC=AB\times \cos B=10\times \cos 30^{\circ}=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}$ b) Diện tích tam giác ABC là: $S_{ABC}=\frac{1}{2}\times AC\times BC=\frac{1}{2}\times 5\times 5\sqrt{3}=\frac{25\sqrt{3}}{2}$ Đáp số: a) $\widehat{B}=30^{\circ}; AC=5; BC=5\sqrt{3}$ b) $S_{ABC}=\frac{25\sqrt{3}}{2}$ a) Ta có $\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$ (tổng 3 góc trong 1 tam giác) $\Rightarrow \widehat{A}=180^{\circ}-\widehat{B}-\widehat{C}=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$ Ta có $\frac{AN}{AC}=\sin C$ $\Rightarrow AN=AC\times \sin C=16\times \sin 30^{\circ}=16\times \frac{1}{2}=8$ Ta có $\frac{BC}{AC}=\cos C$ $\Rightarrow BC=AC\times \cos C=16\times \cos 30^{\circ}=16\times \frac{\sqrt{3}}{2}=8\sqrt{3}$ b) Diện tích tam giác ABC là: $S_{ABC}=\frac{1}{2}\times AN\times BC=\frac{1}{2}\times 8\times 8\sqrt{3}=32\sqrt{3}$ Đáp số: a) $\widehat{A}=60^{\circ}; AN=8; BC=8\sqrt{3}$ b) $S_{ABC}=32\sqrt{3}$