I have this, what seems basic triangle problem, and not certain if the given information is sufficient to solve the problem of finding an angle. We have one main isoceles and another one inside of it. I have attached here a diagram, and we wish to find the angle in red:
The labels blue means : lengths PQ = QR The label green means : lengths QS = QT The given angle PQS = 24 (deg)
We wish to find angle in red , angle RST.
I am trying to figure this out only using geometric principle without trying to use a system of algebraic equations. I tried to create parallel lines to help, but it did not, I tried to make use of principle of exterior angle theorem, but still could not get it. Hope someone here can help me on this.
Although the solution of @Donald Splutterwit might seem like algebra and equations, it is really straight-forward geometry, using the principle that the three angles of a triangle sum to $180^o$.
Perhaps the following may be another helpful way to exhibit the geometry of the situation.
Since $PRQ$ and $STQ$ are isosceles and share vertex $Q$, let the circle with center $Q$ and radius $QT$ intersect $PQ$ at $U$ and $PR$ at $V$, and join $TU$. Thus, since$$\frac{QU}{QP}=\frac{QT}{QR}$$then$$TU\parallel RP$$and arcs$$\ VT= SU$$so that$$\angle VST=\angle STU$$But$$\angle SQU=2\angle STU$$Therefore$$\angle SQU=2\angle VST$$that is$$\angle SQP=2\angle RST$$Thus under the given conditions, the angle we are looking for is always half the given angle.