Предмет: Геометрия, автор: lera13200967

Один из острых углов прямоугольного треугольника на 46° больше другого. Найдите эти углы.​

Ответы

Автор ответа: Alex1567X
2

Ответ:

1) Так как ∆ ABC прямоугольный с прямым углом A => <A=<B+<C=90°

2)<B > < C на 46° => <B=X+46° <C=X

3) <B+<C=90°

X+46°+X=90°

2X=90°-46°

2X=44°

X=22° - угол C

4) <B=90°-<C=58°

Ответ: 90°,22°,58°.

Объяснение:

=> следовательно

< угол

Автор ответа: Пеппер
0

Ответ:

22°    68°

Объяснение:

Сумма острых углов прямоугольного треугольника составляет 90°.

Пусть ∠1=х°,  ∠2=х+46°, тогда

х+х+46=90;  2х=44;  х=22;  ∠1=22°,  ∠2=22+46=68°

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