Explicație pas cu pas:
3)
a) teorema cosinusului:
[tex]BC² = AB² + AC² - 2\cdot AB\cdot AC\cdot \cos(A) \\ = {10}^{2} + {4}^{2} - 2\cdot 10\cdot 4\cdot \cos(60) \\ = 100 + 16 - 80\cdot \frac{1}{2} = 76[/tex]
[tex]=> BC = 2 \sqrt{19} \: cm[/tex]
b) Aria triunghiului:
[tex]Aria = \frac{AB\cdot AC\cdot \sin(A)}{2} \\ = \frac{10\cdot 4\cdot \sin(60) }{2} = 20\cdot \frac{\sqrt{3} }{2} = 10 \sqrt{3} \: {cm}^{2} [/tex]
c) teorema sinusurilor:
[tex]\frac{BC}{ \sin(A) } = \frac{AC}{ \sin(B) } = \frac{AB}{ \sin(C) } = 2R \\ [/tex]
[tex]\frac{2 \sqrt{19} }{ \sin(60) } = \frac{4}{ \sin(B) } = \frac{10}{ \sin(C) } = 2R \\ = > \frac{4\sqrt{19} }{ \sqrt{3} } = \frac{4}{ \sin(B) } = \frac{10}{ \sin(C) } = 2R \\[/tex]
[tex]\frac{4\sqrt{19} }{ \sqrt{3} } = \frac{4}{ \sin(B) } <=> \sin(B) = \frac{\sqrt{3} }{ \sqrt{19} } \\ = > \sin(B) = \frac{ \sqrt{57} }{19} [/tex]
[tex]\frac{4\sqrt{19} }{ \sqrt{3} } = \frac{10}{ \sin(C) } <=> \sin(C) = \frac{10 \sqrt{3} }{4 \sqrt{19} } \\ = >\sin(C) = \frac{5 \sqrt{57} }{38} [/tex]
[tex]\frac{4\sqrt{19} }{ \sqrt{3} } = 2R = > R = \frac{2 \sqrt{57} }{3} \\ [/tex]
5)
[tex]\sin^{2} ( \alpha ) + \cos^{2} ( \alpha ) = 1[/tex]
[tex] \sin(x) = \frac{4}{5} \\ [/tex]
[tex]\cos^{2} (x) = 1 - {\left( \frac{4}{5} \right)}^{2} = 1 - \frac{16}{25} = \frac{9}{25} \\ = > \cos(x) = \frac{3}{5} [/tex]
[tex]\tg(x) = \frac{ \sin(x) }{ \cos(x) } = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3} \\ = >\tg(x) = \frac{4}{3} [/tex]
[tex]\ctg(x) = \frac{1}{ \tg(x) } = > \ctg(x) = \frac{3}{4} \\ [/tex]
