Find the angles x and y in each figure. - Vidyadip

Question

Find the angles x and y in each figure.

✓

Answer

Let given triangle be $\triangle A B C$.
Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A C B=\angle A B C=y$
Also, $\angle A C D$ is an exterior angle.
We know that in a triangle an exterior angle and the interior adjacent angle form a linear pair.
$\therefore \angle A C B+\angle A C D=180^{\circ}$$\quad$ [linear pair]
$\Rightarrow \angle A C B+120^{\circ}=180^{\circ} $
$\Rightarrow \angle A C B=180^{\circ}-120^{\circ}=60^{\circ} $
$\therefore \angle A B C=\angle A C B=60^{\circ}$
$\text { In } \triangle A B C, \angle A+\angle B+\angle A C B=180^{\circ}$
[by angle sum property of a triangle]
$\Rightarrow x+y+y =180^{\circ} $
$\Rightarrow x+60^{\circ}+60^{\circ} =180^{\circ} $
$\Rightarrow x =180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the value of $x$ is $60^{\circ}$ and $y$ is $60^{\circ}$.
(ii)

Then, we have $A B=B C$ and $\angle B=90^{\circ}$.
$\therefore \quad \angle A C B=\angle B A C=x$
[since, angle opposite to equal sides are equal]
$\text { In } \triangle A B C, \angle A+\angle B+\angle A C B=180^{\circ}$
[by angle sum property of a triangle]
$\Rightarrow x+90^{\circ}+x =180^{\circ} $
$\Rightarrow 2 x+90^{\circ} =180^{\circ} $
$\Rightarrow 2 x =180^{\circ}-90^{\circ} $
$\Rightarrow x =\frac{90^{\circ}}{2}=45^{\circ}$
Also, $\angle A C D$ is an exterior angles.
By exterior angle property of a triangle,
Exterior angle = Sum of two interior opposite angles
$\Rightarrow y=\angle B A C+\angle B $
$\Rightarrow y=x+90^{\circ} $
$\Rightarrow y=45^{\circ}+90^{\circ}=135^{\circ}$
Hence, the value of $x$ is $45^{\circ}$ and $y$ is $135^{\circ}$.
(iii)

Let given triangle be $\triangle A B C$.
Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A B C=\angle A C B$
[since, angle opposite to equal sides are equal]
$\Rightarrow \angle A C B=x $
$\text { and } \angle B A C=92^{\circ} \quad \text { [vertically opposite angles] }$
In $\triangle A B C$, by angle sum property of a traingle,
$\angle A B C+\angle B A C+\angle A C B =180^{\circ} $
$\Rightarrow x+92^{\circ}+x =180^{\circ} $
$\Rightarrow 2 x+92^{\circ} =180^{\circ} $
$\Rightarrow 2 x =180^{\circ}-92 \Rightarrow 2 x=88^{\circ} $
$\Rightarrow x =\frac{88^{\circ}}{2} \Rightarrow x=44^{\circ}$
Also, $\angle A C D$ is an exterior angle of $\triangle A B C$.
By exterior angle property of a triangle
Exterior angle = Sum of two interior opposite angles
$\Rightarrow y=\angle B A C+\angle A B C $
$\Rightarrow y=92^{\circ}+x$
$\Rightarrow y=92^{\circ}+44^{\circ}=136^{\circ}$
Hence, the value of $x$ is $44^{\circ}$ and $y$ is $136^{\circ}$.

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