Question

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!

Answer 1

Answer:

secA = $\frac{65}{63}$

Step-by-step explanation:

cscA = $\frac{65}{16}$ = $\frac{hypotenuse}{opposite}$

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

adjacent = $\sqrt{65^2-16^2}$ = 63

Then

secA = $\frac{hypotenuse}{adjacent}$ = $\frac{65}{63}$