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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Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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