Question

in $\triangle def$, $sin d=\frac{36}{39}$. what is $cos e$? a. $\frac{15}{39}$ b. $\frac{39}{15}$ c. $\frac{36}{39}$ d. $\frac{15}{36}$

Explanation:

Step1: Recall trigonometric co - function identity

In a right - triangle, if \(\angle D\) and \(\angle E\) are the two non - right angles, then \(\sin D=\cos E\) because \(\angle D+\angle E = 90^{\circ}\) and the co - function identity states that \(\sin\theta=\cos(90^{\circ}-\theta)\).

Step2: Identify the value of \(\cos E\)

Given that \(\sin D=\frac{36}{39}\), by the co - function identity, \(\cos E=\sin D\). So \(\cos E=\frac{36}{39}\).

Answer:

C. \(\frac{36}{39}\)