Диаметры аб и цд окружности пересекаются в точке о, найдите величину угла, а до если угол б о д равен 154 °

**Решение:** Given: - Angle $б$ о $д$ = 154° To find: - Angle $а$ до Since the diameters $аб$ and $цд$ intersect at point $о$ (center of the circle), we know that angle inscribed in a semicircle is a right angle. Therefore, angle $б$ о $д$ = 90°. Given angle $б$ о $д$ = 154°, and it is a supplementary angle to a right angle in a circle. Supplementary angles add up to $180°$. So, to find angle $а$ до, we can subtract angle $б$ о $д$ from $180°$, since they form a linear pair. Therefore, Angle $а$ до = $180° - 154°$ Angle $а$ до = $26°$ So, the measure of angle $а$ до is $26°$.