Mathematics

In the adjoining figure, it is given that $$ CE \parallel BA , \angle BAC = 80^o $$ and $$ \angle ECD= 35^o $$.Find (i) $$ \angle ACE $$, (ii) $$ \angle ACB $$, (iii) $$ \angle ABC$$.


SOLUTION
From the question,
$$ CE \parallel BA, \angle BAC=80^o , \angle ECD=35^o $$
Now,
(i) $$ \angle BAC= \angle ACE= 80^o $$    ....[$$ \therefore $$ Alternate angles]
(ii) $$ \angle ACB $$,
$$= \angle ACB + \angle ACD = 180^o $$     ....[$$ \therefore $$ Linear pair]
$$ = \angle ACB + \angle ACE + \angle ECD =180^o $$
$$ = \angle ACB + 80 + 35 = 180 $$
$$ = \angle ACB + 125=180 $$
$$ = \angle ACB = 180-115 $$
$$ = \angle ACB=65^o $$
(iii) $$ \angle ABC $$
Let us consider $$ \triangle ABC, $$
 $$ \angle ABC + \angle ACB + \angle BAC = 180^o $$
$$ = \angle ABC + 65+ 80=180^o $$
$$ = \angle ABC+145=180 $$
$$=\angle ABC = 180- 145 $$
$$= \angle ABC=35^o $$


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