I'm having some difficulties trying to solve this problem seen in the picture below. I just don't know where to begin. Any help would be very appreciated, thanks.
Calculate the angel X in the circle
Hello, gingerale!
Plato is absolutely correct.
Calculate the angel X in the circle.
We have cyclic quadrilateral $\displaystyle ABCD.$Code:B o * * * * * * * o C A o * * + * * * * * * * D o * * * *
Draw chords $\displaystyle AB, BC, CD, DA.$
$\displaystyle \text{We have: }\:\begin{Bmatrix}\angle A &=& \frac{1}{2}\overarc{BCD} \\ \angle C &=& \frac{1}{2}\overarc{DAB} \end{Bmatrix}$
$\displaystyle \text{Hence: }\:\angle A + \angle C \;=\;\tfrac{1}{2}\overarc{BCD} + \tfrac{1}{2}\overarc{DAB} \;=\; \tfrac{1}{2}(\overarc{BCD} + \overarc{DAB}) \;=\; \tfrac{1}{2}(360^o)$
$\displaystyle \text{Therefore: }\:\angle A + \angle C \;=\;180^o$
$\displaystyle \text{Opposite angles of a cyclic quadrilateral are supplementary.}$
Generally speaking it is true that the measure of an angle, with vertex on a circle, is half the angle measure of the arc it subtends. Here, the angle opposite x has measure 58 degrees and so subtends an arc of angle 116 degrees. Then entire circle has 360 degrees so that leaves 360- 116= 144 degrees for the arc subtended by angle x.