Question

Se da triunghiul ABC,unde b=√2,c=√3,iar masura unghiului C este egal cu 60°.Sa se afle latura a.

Answer 1

$\displaystyle\\ \text{Vezi desenul atasat.}\\\\ \text{Aplicam teorema sinusurilor:}\\\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ \sin B=\frac{b\sin C}{c}= \frac{\sqrt{2} \cdot \dfrac{\sqrt{3}}{2}}{\sqrt{3}}= \frac{\sqrt{2} \cdot \sqrt{3}}{2\sqrt{3}}=\frac{\sqrt{2}}{2}\\\\ \Longrightarrow~~B = 45^o\\\\ A = 180-60-45 = 75^o$

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$\displaystyle\\ \sin 75 = \sin(30+45)=\sin 30\cos45+\cos 30 \sin 45 =\\\\ =\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2} =\frac{\sqrt{2}}{4}+ \frac{\sqrt{3}\cdot \sqrt{2}}{4}=\boxed{\frac{\sqrt{2}+\sqrt{6}}{4}}\\\\ \text{Rescriem teorema sinusurilor:}$

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$\displaystyle\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ a=\frac{b\sin A}{\sin B}=\frac{\sqrt{2}\cdot \dfrac{\sqrt{2}+\sqrt{6}}{4}}{\dfrac{\sqrt{2}}{2}}=\\\\ =\sqrt{2}\cdot \dfrac{\sqrt{2}+\sqrt{6}}{4}\cdot \dfrac{2}{\sqrt{2}}= \boxed{\bf \frac{\sqrt{2}+\sqrt{6}}{2}~cm } \\\\ \text{Concluzie:}\\ A= 75^o\\ B=45^o\\ C=60^o\\\\ a = \frac{\sqrt{2}+\sqrt{6}}{2}~cm\\\\ b= \sqrt{2}~cm\\ c= \sqrt{3}~cm$

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