Question
In Fig. $\text{AB}=\text{AC}$ and $\angle\text{ACD}=105^\circ,$ find $\angle\text{BAC}.$
✓
Answer
We have, $\text{AB}=\text{AC}$ and $\angle\text{ACD}=(10)5^\circ$
Since, $\angle\text{BCD}=180^\circ=\text{Straight angle}$
$\angle\text{BCA}+\angle\text{ACD}=180^\circ$
$\angle\text{BCA}+(10)5^\circ=180^\circ$
$\angle\text{BCA}=180^\circ-(10)5^\circ$
$\angle\text{BCA}=75^\circ$ And also, $\triangle\text{ABC}$ is an isosceles triangle [AB = AC]
$\angle\text{ABC}=\angle\text{ACB}$ [Angles opposite to equal sides are equal] From $(i)$,
we have $\angle\text{ACB}=75^\circ$
$\angle\text{ABC}=\angle\text{ACB}=75^\circ$ And
also, Sum of Interior angles of a triangle $= 180^\circ$
$\angle\text{ABC}=\angle\text{BCA}+\angle\text{CAB}=180^\circ$
$75^\circ+75^\circ+\angle\text{CAB}=180^\circ$
$150^\circ+\angle\text{BAC}=180^\circ$
$\angle\text{BAC}=180^\circ-150^\circ=30^\circ$
$\angle\text{BAC}=30^\circ$
Since, $\angle\text{BCD}=180^\circ=\text{Straight angle}$
$\angle\text{BCA}+\angle\text{ACD}=180^\circ$
$\angle\text{BCA}+(10)5^\circ=180^\circ$
$\angle\text{BCA}=180^\circ-(10)5^\circ$
$\angle\text{BCA}=75^\circ$ And also, $\triangle\text{ABC}$ is an isosceles triangle [AB = AC]
$\angle\text{ABC}=\angle\text{ACB}$ [Angles opposite to equal sides are equal] From $(i)$,
we have $\angle\text{ACB}=75^\circ$
$\angle\text{ABC}=\angle\text{ACB}=75^\circ$ And
also, Sum of Interior angles of a triangle $= 180^\circ$
$\angle\text{ABC}=\angle\text{BCA}+\angle\text{CAB}=180^\circ$
$75^\circ+75^\circ+\angle\text{CAB}=180^\circ$
$150^\circ+\angle\text{BAC}=180^\circ$
$\angle\text{BAC}=180^\circ-150^\circ=30^\circ$
$\angle\text{BAC}=30^\circ$
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