Odpowiedź:
a)
x - y + √3 = 0
y = x + √3
a = 1 = tg α ⇒ α = 45°
======================
b ) √3 x -3 y + 6 = 0
3 y = √3 x + 6 / : 3
y = [tex]\frac{\sqrt{3} }{3} x + 2[/tex]
a = tg α = [tex]\frac{\sqrt{3} }{3}[/tex] ⇒ α = 30°
===========================
c ) 3 x - √3 y - 2 = 0
√3 y = 3 x - 2 / : √3
y = √3 x - [tex]\frac{2}{\sqrt{3} }[/tex]
a = [tex]\sqrt{3} =[/tex] tg α ⇒ α = 60°
d )
2*sin 60°* x - y + tg 45° = 0
2*[tex]\frac{\sqrt{3} }{2}[/tex] x - y + 1 = 0
√3 x - y + 1 = 0
y = √3 x + 1
a = √3 = tg α ⇒ α = 60°
========================
Szczegółowe wyjaśnienie:
y = a x + b
a = tg α
α - miara kąta nachylenia prostej o równaniu y = a x + b
do osi OX
[tex]\huge\boxed{a) \ \alpha = 45^{o}}\\\\\\\huge\boxed{b) \ \alpha = 30^{o} }\\\\\\\huge\boxed{c) \ \alpha = 60^{o}}\\\\\\\huge\boxed{d) \ \alpha = 60^{o}}[/tex]
Korzystamy z równania kierunkowego prostej:
y = ax + b
gdzie:
a - współczynnik kierunkowy prostej, równy tangensowi kąta nachylenia prostej do osi Ox
[tex]x - y + \sqrt{3} = 0[/tex]
Sprowadzamy równanie ogólne prostej do postaci kierunkowej
[tex]y = x + \sqrt{3}\\\\a = 1 \ \ \rightarrow \ \ tg\alpha = 1 \ \ \rightarrow \ \ \boxed{\alpha = 45^{o}}[/tex]
b)
[tex]\sqrt{3}x - 3y + 6 = 0\\\\3y = \sqrt{3}x + 6 \ \ \ |:3\\\\y = \frac{\sqrt{3}}{3}x +2\\\\a = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ tg\alpha = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ \boxed{\alpha = 30^{o}}[/tex]
c)
[tex]3x - \sqrt{3}y - 2 = 0\\\\\sqrt{3}y =3x-2 \ \ \ |:\sqrt{3}\\\\y = \frac{3}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\cdot x - \frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\\\y = \sqrt{3}x - \frac{2\sqrt{3}}{3}\\\\a = \sqrt{3} \ \ \rightarrow \ \ tg\alpha = \sqrt{3} \ \ \rightarrow \ \ \boxed{\alpha=60^{o}}[/tex]
d)
[tex]2sin60^{o}\cdot x - y + tg45^{o} = 0\\\\2\cdot\frac{\sqrt{3}}{2}\cdot x - y + 1 =0\\\\\sqrt{3}x - y + 1 = 0\\\\y = \sqrt{3}x + 1\\\\a = \sqrt{3} \ \ \rightarrow \ \ tg\alpha = \sqrt{3} \ \ \rightarrow \ \ \boxed{\alpha = 60^{o}}[/tex]
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Odpowiedź:
a)
x - y + √3 = 0
y = x + √3
a = 1 = tg α ⇒ α = 45°
======================
b ) √3 x -3 y + 6 = 0
3 y = √3 x + 6 / : 3
y = [tex]\frac{\sqrt{3} }{3} x + 2[/tex]
a = tg α = [tex]\frac{\sqrt{3} }{3}[/tex] ⇒ α = 30°
===========================
c ) 3 x - √3 y - 2 = 0
√3 y = 3 x - 2 / : √3
y = √3 x - [tex]\frac{2}{\sqrt{3} }[/tex]
a = [tex]\sqrt{3} =[/tex] tg α ⇒ α = 60°
===========================
d )
2*sin 60°* x - y + tg 45° = 0
2*[tex]\frac{\sqrt{3} }{2}[/tex] x - y + 1 = 0
√3 x - y + 1 = 0
y = √3 x + 1
a = √3 = tg α ⇒ α = 60°
========================
Szczegółowe wyjaśnienie:
y = a x + b
a = tg α
α - miara kąta nachylenia prostej o równaniu y = a x + b
do osi OX
Verified answer
Odpowiedź:
[tex]\huge\boxed{a) \ \alpha = 45^{o}}\\\\\\\huge\boxed{b) \ \alpha = 30^{o} }\\\\\\\huge\boxed{c) \ \alpha = 60^{o}}\\\\\\\huge\boxed{d) \ \alpha = 60^{o}}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy z równania kierunkowego prostej:
y = ax + b
gdzie:
a - współczynnik kierunkowy prostej, równy tangensowi kąta nachylenia prostej do osi Ox
a)
[tex]x - y + \sqrt{3} = 0[/tex]
Sprowadzamy równanie ogólne prostej do postaci kierunkowej
[tex]y = x + \sqrt{3}\\\\a = 1 \ \ \rightarrow \ \ tg\alpha = 1 \ \ \rightarrow \ \ \boxed{\alpha = 45^{o}}[/tex]
b)
[tex]\sqrt{3}x - 3y + 6 = 0\\\\3y = \sqrt{3}x + 6 \ \ \ |:3\\\\y = \frac{\sqrt{3}}{3}x +2\\\\a = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ tg\alpha = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ \boxed{\alpha = 30^{o}}[/tex]
c)
[tex]3x - \sqrt{3}y - 2 = 0\\\\\sqrt{3}y =3x-2 \ \ \ |:\sqrt{3}\\\\y = \frac{3}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\cdot x - \frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\\\y = \sqrt{3}x - \frac{2\sqrt{3}}{3}\\\\a = \sqrt{3} \ \ \rightarrow \ \ tg\alpha = \sqrt{3} \ \ \rightarrow \ \ \boxed{\alpha=60^{o}}[/tex]
d)
[tex]2sin60^{o}\cdot x - y + tg45^{o} = 0\\\\2\cdot\frac{\sqrt{3}}{2}\cdot x - y + 1 =0\\\\\sqrt{3}x - y + 1 = 0\\\\y = \sqrt{3}x + 1\\\\a = \sqrt{3} \ \ \rightarrow \ \ tg\alpha = \sqrt{3} \ \ \rightarrow \ \ \boxed{\alpha = 60^{o}}[/tex]