In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 9$ and $BD = 4$.  Find $\sin B$.
The triangle is shown below:

[asy]
pair B,C,D;
C = (0,0);
D = (sqrt(65),0);
B = (sqrt(65),4);
draw(B--C--D--B);
draw(rightanglemark(B,D,C,13));
label("$C$",C,SW);
label("$B$",B,NE);
label("$D$",D,SE);
label("$9$",(B+C)/2,NW);
label("$4$",(B+D)/2,E);
[/asy]

The Pythagorean Theorem gives us $CD = \sqrt{BC^2 - BD^2} = \sqrt{81 - 16} = \sqrt{65}$, so $\sin B = \frac{CD}{BC} = \boxed{\frac{\sqrt{65}}{9}}$.
\frac{\sqrt{65}}{9}
Level 1
Precalculus
Problem: In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 9$ and $BD = 4$.  Find $\sin B$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\sin B$, we need to determine the lengths of the sides of triangle $BCD$. We know that in a right triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the hypotenuse. 

First, we identify the sides relative to angle $B$. The side opposite angle $B$ is $CD$, and the hypotenuse is $BC$. We already have the length of $BC = 9$. 

Next, we need to find the length of side $CD$. We can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Thus, we have:

$BC^2 = BD^2 + CD^2$.

Substituting the known values:

$9^2 = 4^2 + CD^2$.

Calculating the squares gives:

$81 = 16 + CD^2$.

To isolate $CD^2$, we subtract 16 from both sides:

$CD^2 = 81 - 16 = 65$.

Now, we take the square root to find $CD$:

$CD = \sqrt{65}$.

Now we have both the opposite side $CD = \sqrt{65}$ and the hypotenuse $BC = 9$. 

We can now calculate $\sin B$:

$\sin B = \frac{CD}{BC} = \frac{\sqrt{65}}{9}$.

Thus, the final answer is $\frac{\sqrt{65}}{9}$.<|END_OF_LATENT|> The final answer is $\frac{\sqrt{65}}{9}$.