Пошаговое объяснение:
There is bath in the bathroom
There is mirror in the bathroom
There is armchair in the hall
There is carpet in the hall
There is sofa in the hall
There are curtains in the hall
There are cushions in the hall
There is painting in the hall
There is cooker in the kitchen
There is chair in the kitchen
There is table in the kitchen
There is sink in the kitchen
There is fridge in the kitchen
There are pillows in the bedroom
There is desk in the bedroom
There is poster in the bedroom
There is wardrobe in the bedroom
There is bookcase in the bedroom
There are stairs leading from the first to the second floor
Відповідь:
Покрокове пояснення:
3.
△САМ прямоугольний, /_С=30°→ АМ=1/2 МС=4
Из определения синуса и △МАВ → sin/_MBA = MA/MB= 4/4√2=1/√2 → /_MBA=45°
4.
АС=АВ, МА- общая → прямоугольние △САМ и △ВАМ равни →МВ=МС=4см
△САВ равнобедренний →/_АВС=30° → ВА/sin30°=CB/sin120° →BA=6×2/√3×1/2=6/√3
Из прямоугольного △МАВ и из определения косинуса
cos/_MBA=AB/BM=6/√3 :4=√3/2 →/_ МВА=30°
5.
Найдем диагональ квадрата АВ, пусть а равна стороне квадрата
АВ=а√2, по теореме Пифагора
Тогда из △МАВ и определения тангенса имеем
tg/_ABM= MA/AB=a/a√2=1/√2
/_АВМ=actg(1/√2)
Пошаговое объяснение:
There is bath in the bathroom
There is mirror in the bathroom
There is armchair in the hall
There is carpet in the hall
There is sofa in the hall
There are curtains in the hall
There are cushions in the hall
There is painting in the hall
There is cooker in the kitchen
There is chair in the kitchen
There is table in the kitchen
There is sink in the kitchen
There is fridge in the kitchen
There are pillows in the bedroom
There is desk in the bedroom
There is poster in the bedroom
There is wardrobe in the bedroom
There is bookcase in the bedroom
There are stairs leading from the first to the second floor
Відповідь:
Покрокове пояснення:
3.
△САМ прямоугольний, /_С=30°→ АМ=1/2 МС=4
Из определения синуса и △МАВ → sin/_MBA = MA/MB= 4/4√2=1/√2 → /_MBA=45°
4.
АС=АВ, МА- общая → прямоугольние △САМ и △ВАМ равни →МВ=МС=4см
△САВ равнобедренний →/_АВС=30° → ВА/sin30°=CB/sin120° →BA=6×2/√3×1/2=6/√3
Из прямоугольного △МАВ и из определения косинуса
cos/_MBA=AB/BM=6/√3 :4=√3/2 →/_ МВА=30°
5.
Найдем диагональ квадрата АВ, пусть а равна стороне квадрата
АВ=а√2, по теореме Пифагора
Тогда из △МАВ и определения тангенса имеем
tg/_ABM= MA/AB=a/a√2=1/√2
/_АВМ=actg(1/√2)