In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 4 and QC = 1. Find sin angle PBQ.
Let $X$ be the point on line $AD$ such that $PX \perp AD$ and $Y$ be the point on line $AD$ such that $QY\perp AD$, as shown below.
[asy]
pair A,B,C,D,P,Q,X,Y;
A=(0,0);
B=(0,1);
C=(1,1);
D=(1,0);
P=(0,4/5);
Q=(1,4/5);
X=(0,4/5);
Y=(1,4/5);
draw(A--B--C--D--cycle);
draw(P--X);
draw(Q--Y);
draw(P--Q);
label("$A$",A,SW);
label("$B$",B,W);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$X$",X,E);
label("$Y$",Y,E);
[/asy]
Since $BP=4$ and $PC=1$, we have $BC=BP+PC=4+1=5$. Since $DQ=4$ and $QC=1$, we have $DC=DQ+QC=4+1=5$. Therefore, $BC=CD=5$, so $\triangle BCD$ is an isosceles right triangle.
Since $\triangle BCD$ is an isosceles right triangle, we have $\angle BCD = 90^\circ$. Because $\angle PBX$ and $\angle QBY$ are each complementary to $\angle BCD$, we have $\angle PBX = \angle QBY$ and $\angle XBP = \angle YBQ$. Therefore, $\triangle PBX\sim \triangle QBY$.
Since $\triangle PBX\sim \triangle QBY$, we have $\frac{BX}{BY}=\frac{BP}{BQ}=\frac{PB}{QB}$. Also, $\triangle BXP\sim \triangle BQY$, so $\frac{BX}{BQ}=\frac{BP}{BY}$. Therefore, \[\frac{PB}{QB}=\frac{BX}{BY}=\frac{BP}{BQ}=\frac{PB}{BY}.\] Since $BY\neq 0$, we have $QB=BQ$. Therefore, $\triangle BQP$ is an isosceles triangle, so $\angle PBQ = \angle BQP$.
After constructing the altitude from $P$ to $\overline{BQ}$, we have the following diagram:
[asy]
pair A,B,C,D,P,Q,X,Y;
A=(0,0);
B=(0,1);
C=(1,1);
D=(1,0);
P=(0,4/5);
Q=(1,4/5);
X=(0,4/5);
Y=(1,4/5);
draw(A--B--C--D--cycle);
draw(P--X);
draw(Q--Y);
draw(P--Q);
draw(B--Q--P,dashed);
label("$A$",A,SW);
label("$B$",B,W);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$X$",X,E);
label("$Y$",Y,E);
label("$4$",(P+X)/2,W);
label("$1$",(Q+Y)/2,E);
[/asy]
Since $\triangle PBY \sim \triangle QBX$, we have $\frac{BY}{YQ}=\frac{BP}{PX}$. Since $BQ=BP$ (as shown above), we have $\triangle PBQ\cong \triangle PBX$. Therefore, $\angle PBQ = \angle PBX$, so $\angle PBQ = \angle BXP$.
Finally, since $\triangle PBX\sim \triangle BCD$, we have $\frac{XP}{BC}=\frac{BX}{BD}$. Since $BC=5$ and $BD=5$, this simplifies to $\frac{XP}{5}=\frac{BX}{5}$. Because $XP=BP=4$, we have $BP=BX$, so $\angle PBQ = \angle BXP = \angle PBX=45^\circ$. Therefore, $\sin \angle PBQ = \sin \left(45^\circ\right)=\boxed{\frac{\sqrt2}{2}}$.
To find sin angle PBQ, we need to first find the lengths of PB, BQ, and PQ.
Since BP = 4 and PC = 1 in square ABCD, we can calculate the length of BC as the sum of these two lengths:
BC = BP + PC = 4 + 1 = 5.
Similarly, since DQ = 4 and QC = 1 in square ABCD, we can calculate the length of CD as the sum of these two lengths:
CD = DQ + QC = 4 + 1 = 5.
Since ABCD is a square, BC and CD are equal in length:
BC = CD = 5.
Now, let's find the length of BQ. Since BC = 5 and PQ is a diagonal, we can use the Pythagorean theorem to find BQ:
BQ² = BC² + CQ²
BQ² = 5² + 1²
BQ² = 25 + 1
BQ² = 26.
Taking the square root of both sides, we get:
BQ = √26.
Finally, let's find the length of PQ. Since BC = 5, and PQ is a diagonal, we can use the Pythagorean theorem to find PQ:
PQ² = BC² + BP²
PQ² = 5² + 4²
PQ² = 25 + 16
PQ² = 41.
Taking the square root of both sides, we get:
PQ = √41.
Now, we can use the sine ratio to find sin angle PBQ:
sin angle PBQ = opposite/hypotenuse
sin angle PBQ = BQ/PQ
sin angle PBQ = √26/√41
sin angle PBQ = (√26/√41) * (√41/√41) [Multiplying numerator and denominator by √41 to rationalize]
Simplifying further, we get:
sin angle PBQ = √(26/41).
Therefore, sin angle PBQ in square ABCD is √(26/41).