Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonometric ratios sec thetaθ and sin thetaθ.

Answer:[tex]\sin \theta = \dfrac{5}{13}[/tex] and [tex]\sec \theta = -\dfrac{13}{12}[/tex]Step-by-step explanation:Assume that the terminal side of thetaθ passes through the point (−12,5).In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant. Using Pythagoras theorem:[tex]hypotenuse^2=perpendicular^2+base^2[/tex][tex]hypotenuse^2=(5)^2+(12)^2[/tex][tex]hypotenuse^2=25+144[/tex][tex]hypotenuse^2=169[/tex]Taking square root on both sides.[tex]hypotenuse=13[/tex]In a right angled triangle[tex]\sin \theta = \dfrac{opposite}{hypotenuse}[/tex][tex]\sin \theta = \dfrac{5}{13}[/tex][tex]\sec \theta = \dfrac{hypotenuse}{adjacent}[/tex][tex]\sec \theta = \dfrac{13}{12}[/tex]In second quadrant only sine and cosecant are positive.[tex]\sin \theta = \dfrac{5}{13}[/tex] and [tex]\sec \theta = -\dfrac{13}{12}[/tex]