Given csc(A) = 65/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational denominator . Someone please help me!

Respuesta :

Answer:

secA = [tex]\frac{65}{63}[/tex]

Step-by-step explanation:

cscA = [tex]\frac{65}{16}[/tex] = [tex]\frac{hypotenuse}{opposite}[/tex]

This is a right triangle with hypotenuse = 65, opposite = 16 and

adjacent = [tex]\sqrt{65^2-16^2}[/tex] = 63

Then

secA = [tex]\frac{hypotenuse}{adjacent}[/tex] = [tex]\frac{65}{63}[/tex]