Question 325199
I'm assuming that you want to find the value of {{{tan(theta)}}}


Draw a triangle in the third quadrant (since sine is negative, but tangent is positive) and label the two legs to be 'x' and 8. Since the hypotenuse is 9, this means by the pythagorean theorem, we get {{{x^2+8^2=9^2}}}. Solve for 'x' to get {{{x=sqrt(9^2-8^2)=sqrt(81-64)=sqrt(17)}}}



Now recall that tangent = opposite/adjacent. In this case, the opposite side is 8 units long and the adjacent side is {{{sqrt(17)}}} units long. So {{{tan(theta)=8/sqrt(17)=(8/sqrt(17))(sqrt(17)/sqrt(17))=(8*sqrt(17))/17}}}



Or simply put, {{{tan(theta)=(8*sqrt(17))/17}}}