The exterior angle of Triangle ABC is 105°.If angleB = 70°.,find angle A
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let the angle A be x
exterior angle= sum of two interior angle
105°=70°+x
70°+x=105°
x=105°-70°
x=35°
hence angle A is 35°
exterior angle= sum of two interior angle
105°=70°+x
70°+x=105°
x=105°-70°
x=35°
hence angle A is 35°
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