Предмет: Геометрия, автор: Аноним

ДОПОМОЖІТЬ!!! БУДЬЛАСКА

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Ответы

Автор ответа: NNNLLL54

Ответ: $\bf 75^\circ \ ,\ 60^\circ \ ,\ 45^\circ \ .$

Сумма углов треугольника равна 180° .

Из условия задачи: $\alpha =\dfrac{1}{2}\Big(\beta +\gamma \Big)\ ,\ \beta =\dfrac{1}{3}\Big(\alpha +\gamma \Big)$ или $\gamma =\dfrac{1}{3}\Big(\alpha +\beta \Big)$ .

1 вариант.

$\alpha =\dfrac{1}{2}\Big(\beta +\gamma \Big)\ ,\ \beta =\dfrac{1}{3}\Big(\alpha +\gamma \Big)\ \ \Rightarrow \\\\2\alpha =\beta +\gamma \ \ \ (1)\\\\3\beta =\alpha +\gamma \ \ (2)\\\\\\\beta =2\alpha -\gamma \ \ ,\ \ 3\beta =6\alpha -3\gamma \ \ ,\\\\3\beta =\alpha +\gamma =6\alpha -3\gamma \ \ \ \Rightarrow \ \ \ 5\alpha =4\gamma \ \ ,\ \ \boxed{\alpha =\dfrac{4}{5}\, \gamma }$

Вычтем из равенства (1) равенство (2)

$2\alpha -3\beta =\beta -\alpha \ \ \ \Rightarrow \ \ \ 3\alpha =4\beta \ \ ,\ \ \boxed{\beta =\dfrac{3}{4}\, \alpha =\dfrac{3}{4}\cdot \dfrac{4}{5}\gamma =\dfrac{3}{5}\, \gamma }\\\\\\\alpha +\beta +\gamma =\dfrac{4}{5}\, \gamma +\dfrac{3}{5}\, \gamma +\gamma =\dfrac{12}{5}\, \gamma \\\\\dfrac{12}{5}\, \gamma =180^\circ \ \ \Rightarrow \ \ \ \boldsymbol{\gamma }=\dfrac{180^\circ \cdot 5}{12}=\boxed{\bf 75^\circ }$

$\boldsymbol{\alpha }=\dfrac{4}{5}\cdot 75^\circ =\boxed{\boldsymbol{60^\circ }}\\\\\boldsymbol{\beta }=\dfrac{3}{5}\cdot 75^\circ =\boxed{\boldsymbol{45^\circ }}$

2 вариант.

$\alpha =\dfrac{1}{2} \Big(\beta +\gamma \Big)\ ,\ \ \gamma =\dfrac{1}{3} \Big(\alpha +\beta \Big)\ \ \ \Rightarrow \ \ \ 2\alpha =\beta +\gamma \ ,\ \ 3\gamma =\alpha +\beta$

$\alpha =3\gamma -\beta \ \ \Rightarrow \ \ 2\alpha =6\gamma -2\beta =\beta +\gamma \ \ \to \ \ \ 3\beta =5\gamma \ \ ,\ \ \beta =\dfrac{5}{3}\, \gamma$

$2\alpha -3\gamma =\gamma -\alpha \ \ \to \ \ \ 3\alpha =4\gamma \ \ ,\ \ \alpha =\dfrac{4}{3}\, \gamma$

$\alpha +\beta +\gamma =\dfrac{5}{3}\, \gamma +\dfrac{4}{3}\, \gamma +\gamma =\dfrac{12}{3}\, \gamma =4\gamma \ \ ,\ \ 4\gamma =180^\circ \ \ ,\ \ \boldsymbol{\gamma =45^\circ }$