В треугольнике с вершинами в точках A (4; 5; 0), B (2; 3; 0) и C ...

Question

В треугольнике с вершинами в точках A (4; 5; 0), B (2; 3; 0) и C ...

Viktor

В треугольнике с вершинами в точках A (4; 5; 0), B (2; 3; 0) и C (2; 5; 2) найдите в градусах сумму углов при основании AC.

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17.12.2022

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Ответ

Answer 1 · 17.12.2022

Решаю векторами.

Компоненты векторов

{AB}[/tex],

overrightarrow{AC}" title="overrightarrow{AC}" alt="overrightarrow{AC}" /> и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" title="overrightarrow{AB}" title="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" title="overrightarrow{AB}" alt="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" title="overrightarrow{AB}" />,

overrightarrow{AC}" title="overrightarrow{AC}" alt="overrightarrow{AC}" /> и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" alt="overrightarrow{AB}" title="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" alt="overrightarrow{AB}" alt="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" alt="overrightarrow{AB}" />,

overrightarrow{AC}" title="overrightarrow{AC}" alt="overrightarrow{AC}" /> и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" />

$overrightarrow{AC} = textbf{r}_C - textbf{r}_A = { -2; 0; 2 }$

$overrightarrow{CB} = textbf{r}_B - textbf{r}_C = { 0; -2; -2 }$

Нормы (модули, величины, длины) этих векторов:

" title="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" alt="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" />

left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" title="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" alt="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" />

left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" title="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" alt="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" />

Угол между векторами

{AC}" title="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" />

$overrightarrow{AC} = textbf{r}_C - textbf{r}_A = { -2; 0; 2 }$

$overrightarrow{CB} = textbf{r}_B - textbf{r}_C = { 0; -2; -2 }$

Нормы (модули, величины, длины) этих векторов:

" title="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" alt="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" />

left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" title="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" alt="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" />

left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" title="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" alt="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" />

Угол между векторами

{AC}" alt="overrightarrow{AB} = textbf{r}_B - textbf{r}_A = { -2; -2; 0 }" />

$overrightarrow{AC} = textbf{r}_C - textbf{r}_A = { -2; 0; 2 }$

$overrightarrow{CB} = textbf{r}_B - textbf{r}_C = { 0; -2; -2 }$

Нормы (модули, величины, длины) этих векторов:

" title="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" alt="left|overrightarrow{AC}right| = sqrt{(-2)^2+0^2+2^2} = sqrt{8}" />

left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" title="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" alt="left|overrightarrow{AB}right| = sqrt{(-2)^2+(-2)^2+0^2} = sqrt{8}" />

left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" title="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" alt="left|overrightarrow{CB}right| = sqrt{0^2+(-2)^2+(-2)^2} = sqrt{8}" />

Угол между векторами

{AC}" /> и

overrightarrow{AB}" title="overrightarrow{AB}" alt="overrightarrow{AB}" />:

cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" title="overrightarrow{AC}" title="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" title="overrightarrow{AC}" alt="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" title="overrightarrow{AC}" /> и

overrightarrow{AB}" title="overrightarrow{AB}" alt="overrightarrow{AB}" />:

cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" alt="overrightarrow{AC}" title="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" alt="overrightarrow{AC}" alt="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" alt="overrightarrow{AC}" /> и

overrightarrow{AB}" title="overrightarrow{AB}" alt="overrightarrow{AB}" />:

cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" />

Скалярное произведение

$overrightarrow{AC} cdot overrightarrow{AB} = (-2) cdot (-2) + 0 cdot (-2) + 2 cdot 0 = 4$

Имеем

$cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{4}{sqrt{8} cdot sqrt{8}} = frac{4}{8} = frac{1}{2}$

В градусах:

[tex]arccos : frac{1}{2} = 60^{circ}" title="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" />

Скалярное произведение

$overrightarrow{AC} cdot overrightarrow{AB} = (-2) cdot (-2) + 0 cdot (-2) + 2 cdot 0 = 4$

Имеем

$cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{4}{sqrt{8} cdot sqrt{8}} = frac{4}{8} = frac{1}{2}$

В градусах:

[tex]arccos : frac{1}{2} = 60^{circ}" alt="cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{overrightarrow{AC} cdot overrightarrow{AB}}{left|overrightarrow{AC}right| left|overrightarrow{AB}right|}" />

Скалярное произведение

$overrightarrow{AC} cdot overrightarrow{AB} = (-2) cdot (-2) + 0 cdot (-2) + 2 cdot 0 = 4$

Имеем

$cos angle left(overrightarrow{AC}, overrightarrow{AB}right) = frac{4}{sqrt{8} cdot sqrt{8}} = frac{4}{8} = frac{1}{2}$

В градусах:

[tex]arccos : frac{1}{2} = 60^{circ}" />

Угол между векторами

{AC}[/tex] и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" title="overrightarrow{AC}" title="cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" title="overrightarrow{AC}" alt="cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" title="overrightarrow{AC}" /> и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" alt="overrightarrow{AC}" title="cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" alt="overrightarrow{AC}" alt="cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" alt="overrightarrow{AC}" /> и

overrightarrow{CB}" title="overrightarrow{CB}" alt="overrightarrow{CB}" />:

[tex]cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{overrightarrow{AC} cdot overrightarrow{CB}}{left|overrightarrow{AC}right| left|overrightarrow{CB}right|}" />

Скалярное произведение

$overrightarrow{AC} cdot overrightarrow{CB} = (-2) cdot 0 + 0 cdot (-2) + 2 cdot (-2) = -4$

Имеем

$cos angle left(overrightarrow{AC}, overrightarrow{CB}right) = frac{-4}{sqrt{8} cdot sqrt{8}} = frac{-4}{8} = -frac{1}{2}$

В градусах:

$arccos left(-frac{1}{2}right) = 120^{circ}$

Нас интересует угол

$angle left(overrightarrow{CA}, overrightarrow{CB}right) = 180^{circ} - angle left(overrightarrow{AC}, overrightarrow{CB}right) = 180^{circ} - 120^{circ} = 60^{circ}$

Сумма углов будет

+ 60^{circ} = 120^{circ}" title="60^{circ} + 60^{circ} = 120^{circ}" alt="60^{circ} + 60^{circ} = 120^{circ}" />