Question

Помогите пожалуйста решить ,номер 8​

Answer 1

Объяснение:

$A(7;0;1)\ \ \ \ B(2;-5;3)\ \ \ \ C(1;0;0)\\AB*BC=\left|\begin{array}{ccc}\^i&\^j&\^k\\x_B-x_A&y_B-y_A&z_B-z_A\\x_C-x_B&y_C-x_B&z_C-z_B\\\end{array}\right |= \left|\begin{array}{ccc}\^i&\^j&\^k\\2-7&-5-0&3-1&\\1-2&0-(-5)&\ 0-3\\\end{array}\right| =$

$= \left|\begin{array}{ccc}\^i&\^j&\^k\\-5&-5&2\\-1&\ \ 5&-3\\\end{array}\right| =\^i*(-5*(-3)-5*2)-\^j*(-5*(-3)-2*(-1))+$

$+\^k*(-5*5-(-1)*(-5)=5\^i-17\^j-30\^k.$

$S=\frac{1}{2}*|AB*BC|=\frac{1}{2} *\sqrt{5^2+(-17)^2+(-30)^2}=\frac{1}{2}*\sqrt{25+289+900} =\\=\frac{1}{2}* \sqrt{1214} \approx17,42.$

$P=AB+BC+AC\\AB=\sqrt{(2-7)^2+(-5-0)^2+(3-1)^2}=\sqrt{25+25+4}=\sqrt{54} .\\ BC=\sqrt{(1-2)^2+(0-(-5))^2+(0-3)^2}=\sqrt{1+25+9}=\sqrt{35}.\\AC=\sqrt{(1-7)^2+(0-0)^2+(0-1)^2}=\sqrt{36+0+1}=\sqrt{37} . \\P=\sqrt{54}+\sqrt{35} +\sqrt{37}\approx19,35.$

1) ∠ABC.

$AB=(-5;-5;2)\ \ \ \ BC=(-1;5;-3)\ \ \ \ |AB|=\sqrt{54} \ \ \ \ |BC|=\sqrt{35}\\cos_{ABC}=\frac{|-5*(-1)+5*(-5)+2*(-3)|}{\sqrt{54}*\sqrt{35} } =\frac{|5-25-6|}{\sqrt{1890} }=\frac{26}{\sqrt{1890} } \approx=0,6.$

∠ABC≈53°.

2)  ∠BCА.

$BC(-1;5;-3)\ \ \ \ AC(-6;0;-1)\ \ \ \ |BC|=\sqrt{35} \ \ \ \ |AC|=\sqrt{37} \\cos_{BCA}=\frac{|(-1)*(-6)+5*0+(-3)*(-1)|}{\sqrt{35}*\sqrt{37} } =\frac{|6+0+3|}{\sqrt{1295} }=\frac{9}{\sqrt{1295} }\approx0,25.$

∠BCA≈76°.

3)  ∠BАC.

$AB(-5;-5;2)\ \ \ \ AC(-6;0;-1)\ \ \ \ |AB|=\sqrt{54} \ \ \ \ |AC|=\sqrt{37} \\cos_{BAC}=\frac{|-5*(-6)+0*(-5)+2*(-1)|}{\sqrt{54}*\sqrt{37} }=\frac{|30+0-2|}{\sqrt{54*37} } =\frac{28}{\sqrt{1998} }\approx0,626.$

∠BАC≈51°.