1. In ΔABC, ∠C = 54°, the perpendicular bisector of AB at D meets BC at E. If ∠EAC = 42°, then what is the value (in degrees) of ∠ABC?
Write Comment
Comments
By: anil on 05 May 2019 02.26 am
Given : ED is the perpendicular bisectors of AB, $$angle C=54^circ$$ and $$angle EAC=y=42^circ$$
To find : $$angle B=x=?$$ Solution : In $$ riangle$$ EAC, using exterior angle property, => $$angle$$ AEB = $$angle$$ C + $$y$$ => $$angle$$ AEB = $$54^circ+42^circ=96^circ$$ Thus, in $$ riangle$$ AEB, => $$x+z=180^circ-96^circ=84^circ$$ ------------(i) Also, in $$ riangle$$ EAD and $$ riangle$$ BDE AD = DB (DE bisects AB) $$angle$$ EAD = $$angle$$ EDB = $$90^circ$$ DE = DE (Common) Thus, $$ riangle$$ EAD $$cong$$ $$ riangle$$ BDE (By SAS criterion) => $$x=z$$ (By CPCT) Substituting above value in equation (i), we get : => $$x+x=2x=84^circ$$ => $$x=frac{84}{2}=42^circ$$ => Ans - (B)
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
To find : $$angle B=x=?$$ Solution : In $$ riangle$$ EAC, using exterior angle property, => $$angle$$ AEB = $$angle$$ C + $$y$$ => $$angle$$ AEB = $$54^circ+42^circ=96^circ$$ Thus, in $$ riangle$$ AEB, => $$x+z=180^circ-96^circ=84^circ$$ ------------(i) Also, in $$ riangle$$ EAD and $$ riangle$$ BDE AD = DB (DE bisects AB) $$angle$$ EAD = $$angle$$ EDB = $$90^circ$$ DE = DE (Common) Thus, $$ riangle$$ EAD $$cong$$ $$ riangle$$ BDE (By SAS criterion) => $$x=z$$ (By CPCT) Substituting above value in equation (i), we get : => $$x+x=2x=84^circ$$ => $$x=frac{84}{2}=42^circ$$ => Ans - (B)