Question

Given triangle ABC, which equation could be used to find the measure of ∠C?

PLEASE HELP URGENT
right triangle ABC with AB measuring 6, AC measuring 3, and BC measuring 3 square root of 5

cos m∠C = 2 square root of 5 all over 5
sin m∠C = square root of 5 over 5
cos m∠C = square root of 5 over 2
sin m∠C = 2 square root of 5 all over 5

Answer 1

sin m∠C = 2 square root of 5 all over 5

Answer 2

sin m∠C = 2 square root of 5 all over 5

Explanation:

The mnemonic SOH CAH TOA is helpful for sorting this out. A picture can help, too. The mnemonic tells you ...

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

Here, the hypotenuse is 3√5. Putting that in the denominator will result in a factor of (√5)/5 in the sine and cosine values. (This eliminates the third choice.)

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We can examine the ratios that are offered in the answer list and see what they really represent:

(2√5)/5 = AB/BC = sin(C) . . . . not cosine

(√5)/5 = AC/BC = cos(C) . . . . not sine

(√5)/2 = BC/AB = 1/sin(C) = csc(C) . . . . not cosine

(2√5)/5 = AB/BC = sin(C) . . . . . the correct answer choice