Answer:
see explanation
Step-by-step explanation:
sinA = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{BC}{AB}[/tex], thus
sinA = [tex]\frac{2\sqrt{19} }{20}[/tex] = [tex]\frac{\sqrt{19} }{10}[/tex]
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cosA = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{AC}{AB}[/tex], thus
cosA = [tex]\frac{18}{20}[/tex] = [tex]\frac{9}{10}[/tex]
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tanA = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{BC}{AC}[/tex], thus
tanA = [tex]\frac{2\sqrt{19} }{18}[/tex] = [tex]\frac{\sqrt{19} }{9}[/tex]
