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this is a geometry problem meant to be solved using formulation and not construction.

i have tried to solve it a few times but couldn;t solve it.

i searched for the solution, but only got a hint to use thales theoram.

enter image description here

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  • $\begingroup$ if that is a circle with center $A$ then angle $B=78^ \circ$ $\endgroup$
    – Lozenges
    Jun 25, 2017 at 8:13

2 Answers 2

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Let $PQRS$ our quadrilateral, where $PS$ is a diameter of the circle.

$$\measuredangle B=\measuredangle RQB+\measuredangle QRB=\frac{1}{2}\left(\widehat{RS}+\widehat{PQ}\right)=\frac{180^{\circ}-24^{\circ}}{2}=78^{\circ}$$

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  • $\begingroup$ what does $\frac{1}{2}\left(\widehat{RS}+\widehat{PQ}\right)$ mean? $\endgroup$
    – zhirzh
    Jun 25, 2017 at 10:12
  • $\begingroup$ @zhirzh Inscribed angle of the circle equal to the half of the arc, on which the angle rests. $\endgroup$ Jun 25, 2017 at 10:25
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Alternatively: let the inscribed quadrilateral $PQRS$ have the side $PS$ as the diameter of the circle.

Note that $mPRS=90^o$, because it rests on diameter.

Note $mQSR=\frac{1}{2} \cdot mQAR=\frac12 \cdot 24^o=12^o$.

From the right triangle $BRS:$ $$mRBS=180^o-90^o-12^o=78^o.$$

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  • $\begingroup$ why is $mQSR=\frac{1}{2} \dot mQAR$ $\endgroup$
    – zhirzh
    Jun 25, 2017 at 12:43
  • $\begingroup$ Property: inscribed angle is half of central angle when they subtend common arc. $\endgroup$
    – farruhota
    Jun 25, 2017 at 13:01

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