Question

which relationship in the triangle must be true?
a
c

b
c
b

a
○ sin(b) = sin(a)
○ sin(b) = cos(90 − b)
○ cos(b) = sin(180 − b)
○ cos(b) = cos(a)

Explanation:

Step1: Recall trigonometric co-function identity

The co - function identity states that for an acute angle \(\theta\), \(\sin\theta=\cos(90^{\circ}-\theta)\). In a right - triangle \(ABC\) with \(\angle C = 90^{\circ}\), we know that \(\angle A+\angle B=90^{\circ}\), so \(\angle A = 90^{\circ}-\angle B\) and \(\angle B=90^{\circ}-\angle A\).

Let's analyze each option:

Option 1: \(\sin(B)=\sin(A)\)

Since \(\angle A\) and \(\angle B\) are acute angles (in a right - triangle) and \(\angle A
eq\angle B\) (unless the triangle is isosceles right - triangle, but the general case), \(\sin(B)
eq\sin(A)\) in general. For example, if \(\angle A = 30^{\circ}\) and \(\angle B=60^{\circ}\), \(\sin(30^{\circ})=\frac{1}{2}\) and \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\sin(B)
eq\sin(A)\)

Option 2: \(\sin(B)=\cos(90 - B)\)

Using the co - function identity \(\sin\theta=\cos(90^{\circ}-\theta)\), if we let \(\theta = B\), then \(\sin(B)=\cos(90^{\circ}-B)\). Let's verify with an example. If \(B = 30^{\circ}\), then \(\sin(30^{\circ})=\frac{1}{2}\) and \(\cos(90 - 30)^{\circ}=\cos(60^{\circ})=\frac{1}{2}\). If \(B = 60^{\circ}\), \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\) and \(\cos(90 - 60)^{\circ}=\cos(30^{\circ})=\frac{\sqrt{3}}{2}\)

Option 3: \(\cos(B)=\sin(180 - B)\)

We know that \(\sin(180^{\circ}-\theta)=\sin\theta\) (by the sine of supplementary angles formula). So \(\sin(180 - B)=\sin(B)\). And \(\cos(B)
eq\sin(B)\) for non - 45° angles. For example, if \(B = 30^{\circ}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\) and \(\sin(180 - 30)^{\circ}=\sin(150^{\circ})=\frac{1}{2}\), so \(\cos(B)
eq\sin(180 - B)\)

Option 4: \(\cos(B)=\cos(A)\)

Since \(\angle A\) and \(\angle B\) are acute angles and \(\angle A
eq\angle B\) (in general right - triangle), \(\cos(B)
eq\cos(A)\). For example, if \(\angle A = 30^{\circ}\) and \(\angle B = 60^{\circ}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\) and \(\cos(60^{\circ})=\frac{1}{2}\)

Answer:

\(\sin(B)=\cos(90 - B)\) (the second option)