Геометрия, 09.09.2019 15:42, ddasha681


Расстояние между а и в равна 390км. из города а в город в выехал первый автомобиль, а через 3 часа ему на встречу выехал из в со скоростью 70км/ч. найти скорость 1 автомобиля если они встретились на расстоянии 250км от города а

Ответы на вопрос

Ответ разместил: Гость

основание авсd. ас=ав√2=18√2. тогда ав=18

высота so (о-центр основания, точка пересечения диагоналей), ао=1/2ас=9√2

треугольник asb - равнобедренный ан - высота боковой грани,

угол sно - угол наклона бок.грани к плоскости основания

треугольник sно - прямоугольный. угол о = 90 град. угол н=45 град. тогда угол s = 45 град. значит, треугольник - равнобедренный sо=он, он=1/2аd=1/2*18=9.    sн=он√2=9√2


площадь бок. поверхности = 4*s(треугольника аsв)=4*1/2*sн*ав=2*9√2*18=324√2

Ответ разместил: Гость


90: 4=22.5-а


правильно если не я не знаю как по другому!

Ответ разместил: Гость

1) s=1/2bc*sina


2) cos^2a+sin^2a=1


3)ам=20: 2=10

4)по теореме косинусов




Ответ разместил: dariamaniakova

первый автомобиль:


s=250 км

найти v

v=s: t

получаем v=250: 5=10км в час

p.sвремя получили из 3ч через которые выехал второй автомобиль и плюс 2 ч он еще ехал до встречи т.к проехал он 140км со скоростью 70км в час

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