Is it enough to find diameter of a circle using two arbitrary
crossover chords with known length of each partition and angle between this two chords? If it's possible, how?
[Math] Finding diameter of a circle using two chords and angle between them
circles
Related Solutions
Here is a hint:
You have $\dfrac{a}{r} = \sin \alpha$ and $\dfrac{b}{r} = \cos \alpha$.
You may have to take into account that some of these are half what you are interested in.
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 $$\angle BOD=180-2(86.5)=7^\circ$$
Best Answer
If you do not know where they cross (i.e., in what proportions they partition each other), then no.