Question

Rezolvare prin teorema sinusurilor
A=π/6
BC=10 cm
C=π/4
R=?
AB=?

Answer 1

$\it Th.\ sinusurilor\ \Rightarrow \dfrac{AB}{sinC}=\dfrac{BC}{sinA} =2R \ \ \ \ \ \ (*) \\ \\ \\ (*) \Rightarrow \dfrac{AB}{sin\dfrac{\pi}{4}}=\dfrac{10}{sin\dfrac{\pi}{6}} \Rightarrow \dfrac{AB}{\dfrac{\sqrt2}{2}}=\dfrac{10}{\dfrac{1}{2}} \Rightarrow AB=\dfrac{10\cdot\dfrac{\sqrt2}{2}}{\dfrac{1}{2}}=10\sqrt2\ cm$

$\it (*) \Rightarrow \dfrac{10}{\dfrac{1}{2}} =2R \Rightarrow10\cdot2=2R \Rightarrow R = 10\ cm$

Answer 2

$\displaystyle\\\Delta ABC~\text{in care avem: }\\\\A=\frac{\pi}{6}=30^o\\\\BC=10~cm\\\\C=\frac{\pi}{4}=45^o\\\\\text{Se cere: }\\\\R=?~~\text{Raza cercului circumscris}\\AB=? \\\\\text{Rezolvare:}$

$\displaystyle\\\text{Aplicam teorema sinusurilor:}\\\\\frac{BC}{\sin A}=\frac{AB}{\sin C}=\frac{AC}{\sin B}=2R \\\\\frac{10}{\sin 30^o}=\frac{AB}{\sin 45^o}=\frac{AC}{\sin B}=2R \\\\AB = \frac{10\sin 45^o}{\sin 30^o}=\frac{10\cdot \dfrac{\sqrt{2}}{2}}{\dfrac{1}{2}}=\frac{10\sqrt{2}}{2}\cdot\dfrac{2}{1}= \boxed{\bf 10\sqrt{2}~cm}\\\\\\ 2R=\frac{10}{\sin 30^o}\\\\ R==\frac{10}{2\sin 30^o}=\frac{10}{2\cdot\dfrac{1}{2} }= \frac{10}{1}=\boxed{\bf 10~cm}$