Finding the angle between the $2$ radii of a circle Consider a circle with centre $O$. Two chords $AB$ and $CD$ extended
intersect at a point $P$ outside the circle. If $\angle AOC = 43^\circ$ and $\angle BPD =
18^\circ$, then what is the value of $\angle BOD?$
 A: There is an ambiguity in the question in that we do not know whether $A$ and $C$ lie between $B$ and $P$ and $D$ and $P$ respectively, or whether $B$ and $D$ lie between $A$ and $P$ and $C$ and $P$ respectively. Both cases shall be considered.
For where $A$ and $C$ lie between $B$ and $P$ and $D$ and $P$ respectively, we have the figure directly below:-

Note that all triangles formed by joining points on the circle to the centre are isosceles, as their two sides are the radii of the circle. We can easily find angles $\angle OAC=\angle OCA=\frac{1}{2}(180-43)=68.5^\circ$
Considering quadrilateral $ABCD$, the internal angles must sum to $360^\circ$, so we have
$$2(a+b+d)+137=360\Rightarrow a+b+d=111.5 \text{ Eq(1)}$$ 
Examining triangle $BPD$, we have $$a+2b+d+18=180\Rightarrow a+2b+d=162 \text{ Eq(2)}$$
Subtracting $(1)$ from $(2)$ results in $b=50.5$, leading to $$ \angle BOD=180-2(50.5)=79^\circ$$
If we consider the other case, as shown in the figure directly below 

we have the same equation for the quadrilateral $ABCD$, $$2(a+b+d)+137=360\Rightarrow a+b+d=111.5 \text{ Eq(3)}$$
but for triangle $APC$ the equation is
$$a+d+137+18=180\Rightarrow a+d=25 \text{ Eq(4)}$$
so that $b=86.5^\circ$ (subtracting $(4)$ from $(3)$) leading to $$\angle BOD=180-2(86.5)=7^\circ$$
A: Assuming $A$ is between $P$ and $B$, $ C$ is between $P$ and $D$, and I recall my geometry correctly, there is a theorem that the angle at $P$ is half the difference of $\angle BOD$ and $\angle AOC$, which gives $\angle BOD=79^\circ$
