Atrisinājums

Skaidrs, ka \(\sphericalangle CAB=\sphericalangle CBA=80^{\circ}\) un \(\sphericalangle CAM=\sphericalangle BAM=40^{\circ}\). Tā kā \(AM=CM\) (\(M\) uz \(AC\) vidusperpendikula), tad \(\triangle AMC\) - vienādsānu. Tāpēc \(\sphericalangle ACM=\sphericalangle CAM=40^{\circ}\); no šejienes \(\sphericalangle \mathbf{MCB}=\mathbf{40}^{\circ}-\mathbf{20}^{\circ}=\mathbf{20}^{\circ}\).

Novelkam \(BY \perp AD\). Tā kā \(\Delta YAB\) bisektrise ir arī augstums, tad \(\Delta YAB\) - vienādsānu, \(AY=AB\). Tāpēc \(\triangle AYD=\triangle ABD\ (m \ell m)\). Tā kā \(\sphericalangle ADB=180^{\circ}-40^{\circ}-80^{\circ}=60^{\circ}\), tad arī \(\sphericalangle YDA=60^{\circ}\) un \(\sphericalangle YDC=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}\); arī \(\sphericalangle MDC=60^{\circ}\), jo \(\sphericalangle MDC=\sphericalangle ADB\).

Tātad \(\triangle MDC=\Delta YDC\ (\ell m \ell)\), tāpēc \(YD=MD\). Tā kā \(YD=BD\), tad \(MD=BD\), t.i., \(\Delta MDB\) - vienādsānu. Tā kā \(\sphericalangle MDB=180^{\circ}-60^{\circ}=120^{\circ}\), tad \(\sphericalangle \mathbf{MBC}=\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{180^{\circ}}-\mathbf{120^{\circ}}\right)=\mathbf{30^{\circ}}\).