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]
