Question

find the sine of $\angle y$.
$z$
$\sqrt{17}$
$x$
$2\sqrt{19}$
$y$
write your answer in simplified, rationalized form. do not round.
$\sin(y) = $

Explanation:

Step1: Recall sine definition for right triangles

For an acute angle in a right triangle, $\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$

Step2: Identify sides for $\angle Y$

Opposite side to $\angle Y$: $ZX = \sqrt{17}$
Adjacent side to $\angle Y$: $ZY = 2\sqrt{19}$

Step3: Calculate hypotenuse $XY$

Use Pythagorean theorem: $XY = \sqrt{(ZX)^2 + (ZY)^2}$

$$\begin{align*} XY&=\sqrt{(\sqrt{17})^2 + (2\sqrt{19})^2}\\ &=\sqrt{17 + 4\times19}\\ &=\sqrt{17 + 76}\\ &=\sqrt{93} \end{align*}$$

Step4: Compute $\sin(Y)$

Substitute into sine formula: $\sin(Y) = \frac{\sqrt{17}}{\sqrt{93}}$
Rationalize the denominator: $\frac{\sqrt{17}\times\sqrt{93}}{\sqrt{93}\times\sqrt{93}} = \frac{\sqrt{17\times93}}{93} = \frac{\sqrt{1581}}{93}$

Answer:

$\sin(Y) = \frac{\sqrt{1581}}{93}$