Answer:
![SinQ=\frac{15}{17}](https://tex.z-dn.net/?f=SinQ%3D%5Cfrac%7B15%7D%7B17%7D)
Step-by-step explanation:
We draw the right triangle as shown in the image attached.
Δ QRS
Where R is the right angle
Given
QR = 8
QS = 17
Now, using Pythagorean Theorem, we can find RS:
![QR^2+RS^2=QS^2\\8^2+RS^2=17^2\\RS^2=17^2-8^2\\RS^2=225\\RS=\sqrt{225}\\RS=15](https://tex.z-dn.net/?f=QR%5E2%2BRS%5E2%3DQS%5E2%5C%5C8%5E2%2BRS%5E2%3D17%5E2%5C%5CRS%5E2%3D17%5E2-8%5E2%5C%5CRS%5E2%3D225%5C%5CRS%3D%5Csqrt%7B225%7D%5C%5CRS%3D15)
Now, we know the ratio Sine as:
![Sin(\theta)=\frac{Opposite}{Hyotenuse}](https://tex.z-dn.net/?f=Sin%28%5Ctheta%29%3D%5Cfrac%7BOpposite%7D%7BHyotenuse%7D)
Where
is the angle (here angle Q)
Opposite side is RS (which is 15)
Hypotenuse is given as QS (which is 17)
So,
![SinQ=\frac{15}{17}](https://tex.z-dn.net/?f=SinQ%3D%5Cfrac%7B15%7D%7B17%7D)
This is the value of SinQ.
We don't want the angle, so we'll stop here.