Question

for $\triangle abc$, which one is equivalent to $\sin(a)$?
$\bigcirc$ $\cos(b)$
$\bigcirc$ $\sin(b)$
$\bigcirc$ $\cos(c)$
$\bigcirc$ $\tan(c)$

Explanation:

Step1: Use triangle angle sum property

In \(\triangle ABC\), \(A + B + C=180^{\circ}\). Assume \(\triangle ABC\) is a right - triangle with \(C = 90^{\circ}\), then \(A + B=90^{\circ}\), so \(A = 90^{\circ}-B\).

Step2: Apply co - function identity

We know the co - function identity \(\sin(x)=\cos(90^{\circ}-x)\). Substituting \(x = A\), we get \(\sin(A)=\cos(90^{\circ}-A)\). Since \(A + B = 90^{\circ}\) (in right - triangle case), \(\sin(A)=\cos(B)\)

Answer:

\(\cos(B)\)