Предмет: Геометрия, автор: Viktalala

найдите sin, cos, tg острого угла ромба, если Р= 52 см, S=120см

Ответы

Автор ответа: dtnth

Периметр ромба равен 4*сторона

сторона равна периметр4

сторона ромба равна 524=13 см

Площадь ромба равна произведению квадрата стороны на синус угла между сторонами

отсюда синус угла равен площадь робма разделить на квадрат стороны

sin A=120(13^2)=120169

Так как угол А -острый,то cos A=корень(1-sin^2 A)=корень(1-(120169)^2)=

=119169

По одной из основніх формул тригонометрии

tg A=sin Acos A=120169(119169)=120119

Ответ:120169,119169,120119.

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