Given csc(A) = 60/16 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!

Question

contestada

Respuesta :

xKelvin xKelvin · Answer 1 · 2021-06-12T08:52:49+02:00

Answer:

[tex]\displaystyle \sec A=\frac{65}{63}[/tex]

Step-by-step explanation:

We are given that:

[tex]\displaystyle \csc A=\frac{65}{16}[/tex]

Where A is in QI.

And we want to find sec(A).

Recall that cosecant is the ratio of the hypotenuse to the opposite side. So, find the adjacent side using the Pythagorean Theorem:

[tex]a=\sqrt{65^2-16^2}=\sqrt{3969}=63[/tex]

So, with respect to A, our adjacent side is 63, our opposite side is 16, and our hypotenuse is 65.

Since A is in QI, all of our trigonometric ratios will be positive.

Secant is the ratio of the hypotenuse to the adjacent. Hence:

[tex]\displaystyle \sec A=\frac{65}{63}[/tex]

bhadelsimron19 bhadelsimron19 · Answer 2 · 2021-06-12T09:05:50+02:00

Answer:

Step-by-step explanation:

cosec A =60/16

hypotenuse/opposite = 60/16 =15/4 (in simplest form)

therefore hypotenuse = 15 , opposite = 4

then adjacent =? (let be x)

using pythagoras theorem to find adjacent

opposite^2 + adjacent^2 = hypotenuse^2

4^2 + x^2 = 15^2

16 + x^2 = 225

x^2 = 225 - 16

x^2 = 209

[tex]x=\sqrt{209}[/tex]

sec A =hypotenuse/adjacent

[tex]=\frac{15}{\sqrt{209} }[/tex]

[tex]=\frac{15}{\sqrt{209} } * \frac{\sqrt{209} }{\sqrt{209} }[/tex]

=[tex]\frac{15\sqrt{209} }{209}[/tex]