Answer:
sinA = - [tex]\frac{\sqrt{5} }{\sqrt{6} }[/tex]
Step-by-step explanation:
Given
tanA = - [tex]\frac{\sqrt{5} }{1}[/tex] = [tex]\frac{opposite}{adjacent}[/tex]
This is a right triangle with opposite = [tex]\sqrt{5}[/tex], adjacent = 1 and
hypotenuse h² = ([tex]\sqrt{5}[/tex] )² + 1² = 5 + 1 = 6 ⇒ h = [tex]\sqrt{6}[/tex]
Since tanA < 0 and cosA > 0 then A is in the 4th quadrant where sinA < 0
Then
sinA = - [tex]\frac{opposite}{hypotenuse}[/tex] = - [tex]\frac{\sqrt{5} }{\sqrt{6} }[/tex]
