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d: (i) LB(ii) ZEDC(ii) ZADE (iv)(v) ZDEC(vi) ZDCBV110025(vii) LACBVilig, Q. 12 C |
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Answer» i) in triangle BDC Angle B = 180-(100+25) =55ii)angle EDC= DCB (alternate angles) =25iii)angle ADE = 180-( BDC+EDC) = 180-(100+25) = 55iv) in triangle ADE angle AED= 180-(A+ADE) = 180-(55+55)=70V)angle DEC = 180-(ADE) = 180-70 =110vi) angle ECB= AED ( corresponding angle ) angle ECD= 70-25 =45vii)angle ACB= AED ( corresponding angle ) ACB= 70 |
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