Question

Решите пожалуйста заранее спасибо

Answer 1

Ответ:

$1)\ \ AB=12\ ,\ \ \angle B=45^\circ \\\\\dfrac{AC}{AB}=sin45^\circ \ \ \ \to \ \ \ AC=AB\cdot sin45^\circ =12\cdot \dfrac{\sqrt2}{2}=6\sqrt2\\\\\\\dfrac{BC}{AB}=cos45^\circ \ \ \ \to \ \ \ BC=AB\cdot cos45^\circ =12\cdot \dfrac{\sqrt2}{2}=6\sqrt2\\\\\\AC=BC$

$2)\ \ BC=x\ ,\ \ AC=3x\ \ ,\ \ AB=17\\\\x^2+(3x)^2=17^2\ \ \to \ \ 10x^2=289\ \ ,\ \ x^2=\dfrac{289}{10}\ ,\ \ x=\dfrac{17}{\sqrt{10}}=\dfrac{17\sqrt{10}}{10}=1,7\sqrt{10}\\\\BC=1,7\sqrt{10}\ \ ,\ \ AC=3x=5,1\sqrt{10}$

$3)\ \ \angle A=30^\circ \ ,\ \ AC=7\\\\\dfrac{AC}{AB}=cos30^\circ \ \ \to \ \ AB=\dfrac{AC}{cos30^\circ }=\dfrac{7}{\dfrac{\sqrt3}{2}}=\dfrac{14}{\sqrt3}=\dfrac{14\sqrt3}{3}\\\\\\4)\ \ \angle A=60^\circ \ ,\ \ BC=3\sqrt2\\\\\dfrac{BC}{AB}=sin60^\circ \ \ \to \ \ AB=\dfrac{BC}{sin60^\circ }=\dfrac{3\sqrt2}{\dfrac{\sqrt3}{2}}=\sqrt3\cdot 2\sqrt2=2\sqrt6$