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axmid
@axmid
July 2022
1
4
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В треугольнике ABC биссектрисы внешних углов при вершинах B и A пересекаются в точке D.
Найдите угол BDA, если BCA = 28°
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oganesbagoyan
Verified answer
<BDA
=180° -(<DAB+<DBA) = 180° -
(
(<ABC +<ACB)/2) +(<BAC+<BCA)/2 )
) =
180° -
(
(<ABC +<BAC+<BCA)/2+<BCA/2
) =
180° -(90° +<BCA/2 ) =
=
90° - <BCA/2
= 90° - 28°/2 = 90° - 14°
=76°
.
7 votes
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Verified answer
<BDA =180° -(<DAB+<DBA) = 180° -((<ABC +<ACB)/2) +(<BAC+<BCA)/2 ) ) =180° -((<ABC +<BAC+<BCA)/2+<BCA/2 ) =180° -(90° +<BCA/2 ) =
=90° - <BCA/2= 90° - 28°/2 = 90° - 14°=76° .