csc θ = 13/5
[tex]\frac{1}{sin. theta} = \frac{13}{5}[/tex]
When you flip both sides:
sin θ = 5/13
If you know the Pythagorean triples, it will be 5, 12, and 13. Otherwise you could do a² + b² = c²
You can use:
[tex]S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}[/tex]
which is
[tex]Sin \frac{opposite}{hypotenuse} Cos\frac{adjacent}{hypotenuse} Tan\frac{opposite}{adjacent}[/tex]
Or you could draw a triangle, either way works.
To find tangent, you know 13 is the hypotenuse because it is the longest side, and 5 is "opposite" so 12 is "adjacent. So:
tan theta = 5/12
And cosine is adjacent/ hypotenuse, so it is:
cos theta = 12/13
sec theta is the opposite of cos theta, so it is:
sec theta = 13/12
B and C are correct
