Высота конуса равна 13см,а диаметр его основания 26см. найдите угол при вершине осевого сечения конуса. !
anonymus98
2
04.10.2019 20:30
2
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1)Рассмотрим △ АВО
АО=АС/2=26/2=13 см.
Т.к. АО=ВО => △ АВО - р/б => ∡ВАО = ∡АВО = (180° - ∡АОВ)/2 = (180°-90°)/2 = 45° - ∡АВО
2)∡АВС = 2*∡АВО = 2*45° = 90°
ответ: 90°