contestada

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

Respuesta :

sqdancefan

Answer:

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

Step-by-step 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.)

__

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

Ver imagen sqdancefan
ATal

Answer:

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

Otras preguntas