\alpha - kąty przy podstawie
\beta - kąt wierzchołka
\left \{ {{\frac{\alpha}{2}+\frac{\beta}{2}+130^{\circ}=180^{\circ}} \atop {2\alpha+\beta=180^{\circ}}} \right.
\left \{ {{\frac{\alpha}{2}+\frac{\beta}{2}=50^{\circ} \ |*2} \atop {2\alpha+\beta=180^{\circ}}} \right.
\left \{ {{\alpha+\beta=100^{\circ}} \atop {2\alpha+\beta=180^{\circ}}} \right.
\left \{ {{\beta=100^{\circ}-\alpha} \atop {2\alpha+\beta=180^{\circ}}} \right.
2\alpha+100^{\circ}-\alpha=180^{\circ}
\alpha=80^{\circ}
\beta=100^{\circ}-\alpha
\beta=100^{\circ}-80^{\circ}
\beta=20^{\circ}
\left \{ {{\alpha=80^{\circ}} \atop {\beta=20^{\circ}}} \right.
Odpowiedź:
Kąty trójkąta mają miary: 80^{\circ}, \ 80^{\circ}, \ 20^{\circ}.