in the given figure triangle ABC is right angled at C and CD perpendicular AB. If angle A = 50°, find angle ACD, ABC, BCD.
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Given :- /_ CAD=50° (/_ means angle)
/_CDA=90°
/_ACD=50+90+x=180°(angle sum property)
=140+x=180
=180-140=x
=x=/_ACD=40°
if /_CDA=90° than /_CDBalso=90°
/_BCD = 40° (CD cuts 1 angle into 2 different angles)
/_CBD=x+90°+40°=180°(A.S.P)
x+130°=180°
x=180°-130°
/_CBD=x=60°
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