larsonjoseph01
contestada

∠A is an acute angle in a right triangle.
Given that cosA=12\13, what is the ratio for sinA?
Enter your answer in the boxes as a fraction in simplest form.

__
__
__

Respuesta :

HP728
5/ 13 is your answer

Cos A = 12/13

[tex] sin^{2}A[/tex] = [tex] 1 - cos^{2}A [/tex]

Plug in the value of cos A = 12/13

[tex] sin^{2}A[/tex] = [tex] 1 - \left ( \frac{12}{13} \right )^{2} [/tex]

[tex] sin^{2}A[/tex] = [tex] 1 - \left ( \frac{144}{169} \right ) [/tex]

[tex] sin^{2}A[/tex] = [tex] \frac{169-144}{169} [/tex]

[tex] sin^{2}A[/tex] = [tex] \frac{25}{169} [/tex]

Sin A = [tex] \pm [/tex] (5/13)

Otras preguntas