KendallHouston
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∠A is an acute angle in a right triangle.



Given that cosA=15/17, what is the ratio for sinA?



Enter your answer in the boxes as a fraction in simplest form

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marcelo1135
Answer: sinA=8/17 to solve for this you must use Pythagorean theorem and solve for the unknown side. Once you do that you can use SohCahToa to find the fraction that you need to solve for the angle.

Answer:    sinA= 8/17

Step-by-step explanation:  since, sin²A+cos²A=1

                        ⇒  sin²A= 1-cos²A

                        ⇒ sin²A= 1- [tex](\frac{15}{17}) ^2[/tex]

                        ⇒ sin²A=  1- [tex]\frac{15^2}{17^2}[/tex]

                        ⇒  sin²A= [tex]\frac{17^2-15^2}{17^2}[/tex]

                        ⇒   sin²A= [tex]\frac{64}{17^2}[/tex]

                         ⇒ sin²A= [tex]\frac{8^2}{17^2}[/tex]

                         ⇒ sin²A= [tex](\frac{8}{17}) ^2[/tex]

                        ⇒  sinA=[tex]\frac{8}{17}[/tex]

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