Answer:
sin A = [tex]\frac{5}{13}[/tex]
Step-by-step explanation:
As per the statement:
∠A is an acute angle in a right triangle.
Given that:
[tex]\cos A = \frac{12}{13}[/tex]
We have to find Sin A:
Using the formula:
[tex]\sin A= \sqrt{1-\cos^2 A}[/tex]
Substitute the given values we have;
[tex]\sin A= \sqrt{1-(\frac{12}{13})^2}[/tex]
⇒[tex]\sin A = \sqrt{1-\frac{144}{169}} = \sqrt{\frac{169-144}{169}}[/tex]
⇒[tex]\sin A = \sqrt{\frac{25}{169}} = \sqrt{\frac{5^2}{13^2}} = \frac{5}{13}[/tex]
Therefore, the value of sin A as a fraction in simplest form is, [tex]\frac{5}{13}[/tex]
