5. In the diagram, $A B$ is a diameter of a circle with centre $O . C$ and $D$ are points on the circle. $O D$ intersects $A C$ at $P, O C$ intersects $B D$ at $Q$, and $A C$ intersects $B D$ at $R$. If $\angle B O Q=60^{\circ}$ and $\angle A P O=100^{\circ}$, calculate the measure of $\angle B Q O$.
5. Solution 1
Since $O$ is the centre of the circle, then $O A=O C$ and $O B=O D$.
This means that each of $\triangle O A C$ and $\triangle O B D$ is isosceles.
Therefore, $\angle O A C=\angle O C A$ and $\angle O B D=\angle O D B$.
Since $\angle B O Q=60^{\circ}$, then $\angle A O C=180^{\circ}-\angle B O Q=180^{\circ}-60^{\circ}=120^{\circ}$.
Since the angles in $\triangle O A C$ add to $180^{\circ}$ and $\angle O A C=\angle O C A$, then
$$
\angle O A C=\frac{1}{2}\left(180^{\circ}-\angle A O C\right)=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}
$$

Next, consider $\triangle A P O$.
We have $\angle P A O=\angle O A C=30^{\circ}$ and $\angle A P O=100^{\circ}$.
Therefore, $\angle A O P=180^{\circ}-100^{\circ}-30^{\circ}=50^{\circ}$.
Then $\angle B O D=180^{\circ}-\angle A O P=180^{\circ}-50^{\circ}=130^{\circ}$.
Since the angles in $\triangle O B D$ add to $180^{\circ}$ and $\angle O B D=\angle O D B$, then
$$
\angle O B D=\frac{1}{2}\left(180^{\circ}-\angle B O D\right)=\frac{1}{2}\left(180^{\circ}-130^{\circ}\right)=25^{\circ}
$$

Finally, we consider $\triangle B Q O$.
We have $\angle Q B O=\angle O B D=25^{\circ}$ and $\angle B O Q=60^{\circ}$.
Therefore, $\angle B Q O=180^{\circ}-25^{\circ}-60^{\circ}=95^{\circ}$.
Solution 2
Since $O$ is the centre of the circle, then $O A=O C$ and $O B=O D$.
This means that each of $\triangle O A C$ and $\triangle O B D$ is isosceles.
Therefore, $\angle O A C=\angle O C A=x^{\circ}$ and $\angle O B D=\angle O D B=y^{\circ}$ for some numbers $x$ and $y$.
Since $\angle B O Q$ is an exterior angle to $\triangle O A C$, then $\angle O A C+\angle O C A=\angle B O Q$, from which we obtain $x^{\circ}+x^{\circ}=60^{\circ}$ or $2 x=60$ or $x=30$.
Next, consider $\triangle A P O$.
We have $\angle P A O=\angle O A C=30^{\circ}$ and $\angle A P O=100^{\circ}$.
Therefore, $\angle A O P=180^{\circ}-100^{\circ}-30^{\circ}=50^{\circ}$.
Since $\angle A O P$ is an exterior angle to $\triangle O D B$, then $\angle O D B+\angle O B D=\angle A O P$, from which we obtain $y^{\circ}+y^{\circ}=50^{\circ}$ or $2 y=50$ or $y=25$.
Finally, we consider $\triangle B Q O$.
We have $\angle Q B O=\angle O B D=25^{\circ}$ and $\angle B O Q=60^{\circ}$.
Therefore, $\angle B Q O=180^{\circ}-25^{\circ}-60^{\circ}=95^{\circ}$.
Solution 3
Since $O$ is the centre of the circle, then $\angle B O C=2 \angle B A C$. (This uses the property of circles that the angle formed by a chord at the centre of the circle is twice the angle formed on the major arc.) Since $\angle B O C=\angle B O Q=60^{\circ}$, then $\angle B A C=30^{\circ}$.
Since $O$ is the centre of the circle, then $O A=O C$ and $O B=O D$.
This means that each of $\triangle O A C$ and $\triangle O B D$ is isosceles.
Next, consider $\triangle A P O$.
We have $\angle P A O=\angle B A C=30^{\circ}$ and $\angle A P O=100^{\circ}$.
Therefore, $\angle A O P=180^{\circ}-100^{\circ}-30^{\circ}=50^{\circ}$.
Using the same circle property as above, $\angle A B D=\frac{1}{2} \angle A O D=\frac{1}{2} \angle A O P=25^{\circ}$.
Finally, we consider $\triangle B Q O$.
We have $\angle Q B O=\angle A B D=25^{\circ}$ and $\angle B O Q=60^{\circ}$.
Therefore, $\angle B Q O=180^{\circ}-25^{\circ}-60^{\circ}=95^{\circ}$.
ANSWER: $95^{\circ}$