Respuesta :
freq. used Pythagorean triangles
3:4:5 and. 5:12:13
on unit circle, valid values for cos [x-axis] sin [y-axis] are
3/5, 4/5
5/13, 12/13
3:4:5 and. 5:12:13
on unit circle, valid values for cos [x-axis] sin [y-axis] are
3/5, 4/5
5/13, 12/13
Answer:
[tex](\frac{6}{7},(\frac{\sqrt{13}}{7})[/tex] and [tex](\frac{5}{13} ,\frac{12}{13})[/tex] are the only two points that lies on unit circle.
Step-by-step explanation:
The general equation for the circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where (h,k) is the point of center.
So, For a circle with center at (0,0) and radius 1 unit. equation becomes,
[tex](x)^2+(y)^2=1[/tex] .........(1)
Now, from the given points are on unit circle or not , we just have to put the values for x and y and check whether they satisfy the equation of circle.
1) [tex](\frac{1}{3} ,\frac{2}{3})[/tex]
Substitute for x and y in Left side of (1), we get,
[tex](x)^2+(y)^2=1[/tex]
[tex]\Rightarrow (\frac{1}{3})^2+(\frac{2}{3})^2[/tex]
[tex]\Rightarrow (\frac{1}{9})+(\frac{4}{9})[/tex]
[tex]\Rightarrow (\frac{1+4}{9})=\frac{5}{9} \neq 1[/tex]
Thus, [tex](\frac{1}{3} ,\frac{2}{3})[/tex] point do not lies on unit circle.
2) [tex](\frac{4}{3} ,\frac{4}{5})[/tex]
Substitute for x and y in Left side of (1), we get,
[tex](x)^2+(y)^2=1[/tex]
[tex]\Rightarrow (\frac{4}{3})^2+(\frac{4}{5})^2[/tex]
[tex]\Rightarrow (\frac{16}{9})+(\frac{16}{25})[/tex]
[tex]\Rightarrow (\frac{544}{225}) \neq 1[/tex]
Thus, [tex](\frac{4}{3} ,\frac{4}{5})[/tex] point do not lies on unit circle.
3) [tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex]
Substitute for x and y in Left side of (1), we get,
[tex](x)^2+(y)^2=1[/tex]
[tex]\Rightarrow (\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2[/tex]
[tex]\Rightarrow (\frac{36}{49})+(\frac{13}{49})[/tex]
[tex]\Rightarrow (\frac{36+13}{49})=\frac{49}{49}=1[/tex]
Thus,[tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex] point lies on unit circle.
4) [tex](\frac{5}{13} ,\frac{12}{13})[/tex]
Substitute for x and y in Left side of (1), we get,
[tex](x)^2+(y)^2=1[/tex]
[tex]\Rightarrow (\frac{5}{13})^2+(\frac{12}{13})^2[/tex]
[tex]\Rightarrow (\frac{25}{169})+(\frac{144}{169})[/tex]
[tex]\Rightarrow (\frac{25+144}{169})=\frac{169}{169}=1[/tex]
Thus, [tex](\frac{5}{13} ,\frac{12}{13})[/tex] point lies on unit circle.
Thus, only two points lies on unit circle.
[tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex] and [tex](\frac{5}{13} ,\frac{12}{13})[/tex] are the only two points that lies on unit circle.