In the diagram, the six small squares all have side length 2. Lines are drawn from $O$ to $P$ and $O$ to $Q$. The measure of $\angle P O Q$ in degrees, accurate to one decimal place, is
(A) 15.0
(B) 25.5
(C) 26.6
(D) 22.5
(E) 30.0

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-094.jpg?height=306&width=436&top_left_y=660&top_left_x=1297)
We add label $R$ and $S$ in the diagram.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-185.jpg?height=325&width=434&top_left_y=2282&top_left_x=908)

Since the side length of each small square is 2 , then $O R=R P=2(2)=4$ and $\angle O R P=90^{\circ}$.

Since $\triangle O R P$ is isosceles and right-angled, then $\angle R O P=45^{\circ}$.

In $\triangle O S Q$, we have $Q S=2, O S=3(2)=6$ and $\angle O S Q=90^{\circ}$.

Therefore, $\tan (\angle Q O S)=\frac{2}{6}=\frac{1}{3}$, so $\angle Q O S \approx 18.43^{\circ}$.

Thus, $\angle P O Q=\angle P O R-\angle Q O S \approx 45^{\circ}-18.43^{\circ}=26.57^{\circ}$, which, to the nearest tenth of a degree is $26.6^{\circ}$.

ANsweR: $(\mathrm{C})$