How do you solve this problem?

Answer:[tex]sin(A) = \frac{21}{{29} }[/tex] [tex]cos(A)=\frac{20}{29 }[/tex] [tex]tan(A)=\frac{21}{20}[/tex] [tex]sin(C) = \frac{20}{{29} }[/tex] [tex]cos(C)=\frac{21}{29 }[/tex] [tex]tan(C)=\frac{20}{21}[/tex] Step-by-step explanation:So first we need to make sure we know the trig identities to solve this problem.[tex]sin(\alpha )=\frac{opposite}{hypotenuse}[/tex] [tex]cos(\alpha )=\frac{adjacent}{hypotenuse}[/tex] [tex]tan(\alpha )=\frac{opposite}{adjacent}[/tex] Here we only have two legs of the triangle, so we will need to use the Pythagorean Theorem a² + b² = c² to solve for the missing leg, the hypotenuse in this case.Solving for the hypotenuse, c, we get [tex]c = \sqrt{a^{2} +b^{2} }[/tex] Here a = 20 and b = 21, so plugging in these values to the equation we get: [tex]c= \sqrt{(20)^{2}+(21)^{2} } =\sqrt{400+441} =\sqrt{841}=29[/tex] Now we can use the trig identities to figure out the missing values for the problem[tex]sin(A) = \frac{21}{{29} }[/tex] [tex]cos(A)=\frac{20}{29 }[/tex] [tex]tan(A)=\frac{21}{20}[/tex] [tex]sin(C) = \frac{20}{{29} }[/tex] [tex]cos(C)=\frac{21}{29 }[/tex] [tex]tan(C)=\frac{20}{21}[/tex]