Question

How do you solve this problem

Answer 1

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

`cos(A)=(20)/(29 )`

`tan(A)=(21)/(20)`

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

`cos(C)=(21)/(29 )`

`tan(C)=(20)/(21)`

Explanation:

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

• 
  `sin(\alpha )=(opposite)/(hypotenuse)`
• 
  `cos(\alpha )=(adjacent)/(hypotenuse)`
• 
  `tan(\alpha )=(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) } =√(400+441) =√(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)=(20)/(29 )`
• 
  `tan(A)=(21)/(20)`
• 
  `sin(C) = \frac{20}{{29} }`
• 
  `cos(C)=(21)/(29 )`
• 
  `tan(C)=(20)/(21)`