Question

given right triangle def, what is the value of sin(e)? o $\frac{3}{5}$ o $\frac{3}{4}$ o $\frac{4}{5}$ o $\frac{4}{3}$

Explanation:

Step1: Recall sine - ratio definition

In a right - triangle, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\sin(E)$ in right - triangle $DEF$, the side opposite to angle $E$ is $DF$ and the hypotenuse is $EF$.

Step2: Use the Pythagorean theorem to find $DF$

By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = EF = 10$ and $a = DE=8$. Let $DF = b$. Then $b=\sqrt{EF^{2}-DE^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6$.

Step3: Calculate $\sin(E)$

$\sin(E)=\frac{DF}{EF}$. Since $DF = 6$ and $EF = 10$, $\sin(E)=\frac{6}{10}=\frac{3}{5}$.

Answer:

$\frac{3}{5}$