Find angle ACB please soon as possible
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As in triangle ABC , AC= AB , So it is an isosceles. Hence , <B = <C
As , <A+<B+<C =180°, Therefore
80°+ 2<B = 180°
<B = 100°/2= 50°
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In triangle, ACB
AC= AB this shows its a Isosceles traingle
as= Angle A + AngleB+ angleC
80°+ 2 AngleB=180°
AngleB = 100/2= 50°
Thanks..
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