I'm stuck at this problem for hours. This is from a book from Kiselev's geometry.
2026-03-29 07:40:19.1774770019
Into a given circle, inscribe a triangle, given the sum of two of its sides and the angle opposite to one of them.
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By drawing a tangent to the given circle, take a chord which forms with the tangent the given angle. This chord is one of the sides of the given sum. Note that all the inscribed angles intercepting the corresponding arc are congruent to the given angle. Let $P$ and $Q$ be the endpoints of the chord. From $P$ extend the chord outside the circle to obtain a segment $RQ$ of length equal to the given sum. Let $S$ be the intersection of the arc and the circle centred at $P$ of radius $RP$. Then $\triangle PQS$ is the required triangle.