# The terminal side of #theta# lies on the line #y=1/3x# in quadrant III, how do you find the values of the six trigonometric functions by finding a point on the line?

##### 1 Answer

We're in the third quadrant, where both the x-axis and the y-axis is negative.

I would recommend looking for the first coordinate on the line after the origin where both x and y values are integers. This would be

We now have enough information to determine

#tantheta = "opposite"/"adjacent" = -1/(-3) = 1/3#

#cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = (-3)/(-1) = 3#

We must find the hypotenuse of the triangle to find the other ratios.

#(-3)^2 + (-1)^2 = h^2#

#9 + 1= h^2#

#h = sqrt(10)#

We can now apply the definitions of the other ratios to solve.

#sintheta = "opposite"/"hypotenuse" = -1/sqrt(10) = (-sqrt(10))/10#

#costheta = "adjacent"/"hypotenuse"= -3/sqrt(10) = (-3sqrt(10))/10#

#sectheta = "hypotenuse"/"adjacent" = sqrt(10)/-3 = -sqrt(10)/3#

#csctheta = "hypotenuse"/"opposite" = sqrt(10)/-1 = -sqrt(10)#

Hopefully this helps!