Marinada '23

[lang = hr] U unutrašnjosti trokuta $ABC$ u kome je $\angle BAC=60^{\circ}$ i $\angle ABC=20^{\circ}$ uočena je točka $Q$ takva da je $\angle QCB= 3 \cdot \angle QBC$. Simetrale $\angle QBA$ i $\angle QCA$ sijeku se u točki P, pri čemu je $\angle PAB=20^{\circ}$. Nađi $\angle QBC$. Rješenje zaokruži na dvije decimale. [/lang] [lang = en] In the interior of the triangle $ABC$ in which $\angle BAC=60^{\circ}$ and $\angle ABC=20^{\circ}$ the point $Q$ was observed such that $\angle QCB= 3 \cdot \angle QBC$. Bisectors $\angle QBA$ and $\angle QCA$ intersect at point P, where $\angle PAB=20^{\circ}$. Find $\angle QBC$. Round the solution to two decimal places. [/lang]