Question

How do you solve this problem

Answer 1

Answer:

$sin(A) = \frac{21}{{29} }$

$cos(A)=\frac{20}{29 }$

$tan(A)=\frac{21}{20}$

$sin(C) = \frac{20}{{29} }$

$cos(C)=\frac{21}{29 }$

$tan(C)=\frac{20}{21}$

Step-by-step explanation:

So first we need to make sure we know the trig identities to solve this problem.

• $sin(\alpha )=\frac{opposite}{hypotenuse}$
• $cos(\alpha )=\frac{adjacent}{hypotenuse}$
• $tan(\alpha )=\frac{opposite}{adjacent}$

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 $c = \sqrt{a^{2} +b^{2} }$
• Here a = 20 and b = 21, so plugging in these values to the equation we get: $c= \sqrt{(20)^{2}+(21)^{2} } =\sqrt{400+441} =\sqrt{841}=29$

Now we can use the trig identities to figure out the missing values for the problem

• $sin(A) = \frac{21}{{29} }$
• $cos(A)=\frac{20}{29 }$
• $tan(A)=\frac{21}{20}$
• $sin(C) = \frac{20}{{29} }$
• $cos(C)=\frac{21}{29 }$
• $tan(C)=\frac{20}{21}$