Malia tried to prove thacos(θ)=sin(90°−θ)cosine, left parenthesis, theta, Her proof is not correct. Statement Reason
1 m\angle C = 90\degree - \thetam∠C=90°−θm, angle, C, equals, 90, degree, minus, theta The acute angles in a right triangle are complementary.
2 \sin(90\degree-\theta)=\dfrac{AC}{B C}sin(90°−θ)=
BC
AC

sine, left parenthesis, 90, degree, minus, theta, right parenthesis, equals, start fraction, A, C, divided by, B, C, end fraction Definition of sine.
3 \cos(\theta)=\dfrac{AC}{BC}cos(θ)=
BC
AC

A, C, divided by, B, C, end fraction Definition of cosine.
4 cos(θ)=sin(90°−θ)
What is the first mistake in Malia's proof?
Choose 1

(Choice A)
A
The measures of \angle B∠Bangle, B and \angle C∠Cangle, C add up to 180\degree180°180, degree, not 90\degree90°90, degree.

(Choice B)
B
Malia used the wrong sides in her ratio for \sin(90\degree-\theta)sin(90°−θ)sin e, left parenthesis, 90, degree, minus, theta, right parenthesis.

(Choice C)
C
Malia used the wrong sides in her ratio for \cos(\theta)cos(θ)cosine, left parenthesis, theta, right parenthesis.

(Choice D)
D
Malia substituted a value that wasn't equivalent to the one she replaced.