Question

triangle def is right angled at f. angle def = 36\circ and df = 15\text{ cm}. the length of de, in \text{cm}, is given by
a. 15\sin 36\circ
b. \dfrac{15}{\cos 36\circ}
c. 15\cos 54\circ
d. \dfrac{15}{\cos 54\circ}

Explanation:

Step1: Recall sine - ratio in right - triangle

In right - triangle $DEF$ with right - angle at $F$, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = \angle DEF = 36^{\circ}$, the side opposite to $\angle DEF$ is $DF$ and the hypotenuse is $DE$. So, $\sin(\angle DEF)=\sin36^{\circ}=\frac{DF}{DE}$.

Step2: Solve for $DE$

Given $DF = 15$ cm. From $\sin36^{\circ}=\frac{DF}{DE}$, we can cross - multiply to get $DE=\frac{DF}{\sin36^{\circ}}$. Also, note that $\cos(90^{\circ}-\alpha)=\sin\alpha$. Since $90^{\circ}- 36^{\circ}=54^{\circ}$, $\sin36^{\circ}=\cos54^{\circ}$, and $DE=\frac{15}{\cos54^{\circ}}$.

Answer:

D. $\frac{15}{\cos54^{\circ}}$