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The side QR of a triangle PQR is extended to a point S. If $$\angle PRS = 104^\circ$$ and $$\angle RQP = \frac{3}{5} \angle QPR$$ , then the value of $$\angle QPR$$ is:
Using external angle theorem, Angle PRS = Angle RQP + QPR.
Angle RPQ = 3/5 of Angle QPR.
104 = $$\frac{3}{5}QPR\ +\ QPR$$
104 = $$\frac{8}{5}QPR$$
Angle QPR = 65 degrees.
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