Question

find the sine of ∠p.
triangle with right angle at q, pq = 5, qr = √30, pr = √55
write your answer in simplified, rationalized form. do not round.
sin(p) = box with fraction and square root buttons

Explanation:

Step1: Recall sine definition in right triangle

In a right triangle, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle P$, the opposite side to $\angle P$ is $QR = \sqrt{30}$, and the hypotenuse is $PR = \sqrt{55}$.

Step2: Apply the sine formula

So, $\sin(P) = \frac{QR}{PR} = \frac{\sqrt{30}}{\sqrt{55}}$. To rationalize the denominator, multiply numerator and denominator by $\sqrt{55}$: $\frac{\sqrt{30} \times \sqrt{55}}{\sqrt{55} \times \sqrt{55}} = \frac{\sqrt{30 \times 55}}{55} = \frac{\sqrt{1650}}{55}$. Simplify $\sqrt{1650}$: $1650 = 25 \times 66$, so $\sqrt{1650} = 5\sqrt{66}$. Then $\frac{5\sqrt{66}}{55} = \frac{\sqrt{66}}{11}$.

Answer:

$\frac{\sqrt{66}}{11}$