Question
Find the interior angles of the following triangles:
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Answer
In $ABC,$
$AB = AC$
$\Rightarrow \angle ACB = \angle ABC ....(1)($angles opposite two equal sides are equal$)$
Now, $\angle ACB + \angle ACD = 180^\circ ...($linear pair$)$
$\Rightarrow \angle ACB = 180^\circ - \angle ACD$
$\Rightarrow \angle ACB = 180^\circ - 105^\circ $
$\Rightarrow \angle ABC = 75^\circ $
$\Rightarrow \angle ABC = 75^\circ ....[$From $(1)]$
By angle sum property, in $\triangle ABC$
$\angle ABC + \angle ACB - \angle BAC = 180^\circ $
$\Rightarrow 75^\circ + 75^\circ + \angle BAC = 180^\circ $
$\Rightarrow 150^\circ + \angle BAC = 180^\circ $
$\Rightarrow \angle BAC = 180^\circ - 150^\circ $
$\Rightarrow \angle BAC = 30^\circ $
Hence, in $\triangle ABC, \angle A = 30^\circ , \angle B = 75^\circ $ and $\angle C = 75^\circ .$
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