Question

In triangle ABC, angle C is the right angle. Which statement is always true? A. Sin(A)=cos(A)
B. Sin(A)=cos(B)
C. Sin(A)=cos(C)

Answer 1

Given this, we know:
A+B+90° = 180°
Therefore:
90°-A = B and 90°-B = A
We also know:
sin(A) = a/c
cos(B) = a/c

therefore,
Sin (A)=Cos (B)

Answer 2

Given that angle C is a right angle in triangle ABC, the statement from the given options that will always be true considering their trigonometry ratios is:

B. Sin(A) = Cos(B)

Recall:

• Trigonometry ratios are: SOH CAH TOA.
• They can be used when solving a right triangle.

The image of the right triangle ABC is shown in the attachment below.

We are given the following:

AC = 5

AB = 13

BC = 12

Let's find the trigonometry ratios for angle A and angle B respectively.

Trigonometry ratios of angle A as the reference angle:

$Sin(A) = \frac{Opp}{Hyp} = \frac{12}{13}$

$Cos(A) = \frac{Adj}{Hyp} = \frac{5}{13}$

$Tan(A) = \frac{Opp}{Adj} = \frac{12}{5}$

Trigonometry ratios of angle B as the reference angle:

$Sin(B) = \frac{Opp}{Hyp} = \frac{5}{13}$

$Cos(B) = \frac{Adj}{Hyp} = \frac{12}{13}$

$Tan(B) = \frac{Opp}{Adj} = \frac{5}{12}$

From the above, we can deduce the following:

• Sin(A) = Cos(B)
• Cos(A) = Sin(B)

Therefore, given that angle C is a right angle in triangle ABC, the statement from the given options that will always be true considering their trigonometry ratios is:

B. Sin(A) = Cos(B)

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