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solve in an isosceles triangle ABC
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If DE||BC then \[\angle ADE= \angle ABC ( corresponding - angles) -----1\] Therefore \[x+y-36^0 =2x\] \[y-36^0 =2x-x \] \[y-36^0 =x ----(2)\] Now since AB = AC (given) therefore \[\angle ABC= \angle ACB ( angles-opposite -equal-sides)\]Hence 2x=y-2 i.e. 2x-2 =y --------(3) from 2 and 3 we have 2x-2-36^0 =x 2x-x =38^0 x=38^0
just plse check it again its x = 34 degree or 38 degree
It is x=38^0
y=2x-2 = 2*38-2=76-2=70^0
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