Answer:
[tex]\sin(\theta)=\frac{\sqrt{13}}{7}\approx0.5151[/tex]
Step-by-step explanation:
So we know that Θ is acute and that:
[tex]\cos(\theta)=6/7[/tex]
First, note that since Θ is acute, it is between 0 and 90 degrees, meaning that all of our trig ratios will be positive since the angle is in the Quadrant I.
The Pythagorean Identity is:
[tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex]
We know that cosine is 6/7. Substitute:
[tex]\sin^2(\theta)+(\frac{6}{7})^2=1[/tex]
Square:
[tex]\sin^2(\theta)+\frac{36}{49}=1[/tex]
Subtract both sides by 36/49. Change the 1 to 49/49 to make a common denominator:
[tex]\sin^2(\theta)=\frac{49}{49}-\frac{36}{49}[/tex]
Subtract:
[tex]\sin^2(\theta)=\frac{13}{49}[/tex]
Take the square root of both sides:
[tex]\sin(\theta)=\pm\sqrt{\frac{13}{49}}[/tex]
Simplify:
[tex]\sin(\theta)=\pm\frac{\sqrt{13}}{7}[/tex]
As mentioned previously, since Θ is acute, all of the trig functions must be positive. So, we can ignore our negative answer:
[tex]\sin(\theta)=\frac{\sqrt{13}}{7}\approx0.5151[/tex]
and we're done!
