Consider the following star figure. If $\angle p + \angle q + \angle r + \angle s + \angle t = 500^\circ$, find $\angle A + \angle B + \angle C + \angle D + \angle E$.
What I Tried: Here is the figure :-
Some might already know the sum of the interior angles of an $n$-angled star, is $(n - 4)180^\circ$. Using that one might not deduce that fast that the answer is not $180^\circ$ in this case. In fact, the star is defined such that we draw it from a specific point, and continue making it without lifting the pencil.
For this, this is not the case. In fact it is more difficult to answer this question, from my point of view. The answer was given as $140^\circ$ . Joining lines and taking variables will take a long time, and in some cases the extended lines will not meet at good places, making it more hard.
Can anyone help me how is that the answer?
Considering the angle sum of the star figure, which is actually a 10-gon:
$$(360^\circ - p)+\dots + (360^\circ -t)+\angle A + \dots + \angle E = (10-2)\times 180^\circ $$
$$360^\circ \times 5 - (p+q+r+s+t)+\angle A + \dots + \angle E = 8\times 180^\circ $$
Now solve for $\angle A + \dots + \angle E$.