Question
Find angle x in each figure.
Image

Answer

(i)
Image
Let the given triangle be $\triangle A B C$, then we have $A B=A C$. We know that angle opposite to equal sides of an isosceles triangle are equal.
$\therefore \quad x=40^{\circ}$
Hence, the value of $x$ is $40^{\circ}$.
(ii)
Image
Let the given triangle be $\triangle A B C$. Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \angle A=\angle C=45^{\circ}$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
Now, in $\triangle A B C$ by angle sum property of a triangle,
$\angle A+\angle B+\angle C=180^{\circ} \Rightarrow 45^{\circ}+x+45^{\circ}=180^{\circ} $
${[\angle A=\angle B=45 \text { because of equal side}]}$
$\Rightarrow \quad x+90^{\circ}=180^{\circ} \Rightarrow x=180^{\circ}-90^{\circ}=90^{\circ}$
Hence, the value of $x$ is $90^{\circ}$.
(iii) Ans. $x=50^{\circ}$
(iv) Ans. $x=40^{\circ}$
(v) Ans. $45^{\circ}$
(vi) Ans. $x=70^{\circ}$
(vii)
Image
Let the given triangle be $\triangle A B C$.
Then, we have
$A B=A C$
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A C B=\angle B$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
$\angle A C B=x$
We know that linear pair of angles is supplementry.
$\therefore \angle A C B+\angle A C D=180^{\circ} \quad \text { [linear pair] } $
$\Rightarrow x+120^{\circ}=180^{\circ} {[\because \angle A C D=120, \text { given] }} $
$\Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the value of $x$ is $60^{\circ}$.
(viii)
Image
Let the given triangle be $\triangle A B C$. Then, we have
$A B=A C$
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle B=\angle C$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
Also, $\angle D A C$ is an exterior angle of $\triangle A B C$.
Now, by exterior angle property of a triangle,
Sum of two interior opposite angles = Exterior angle
$\Rightarrow x+x=110^{\circ} \Rightarrow 2 x=110^{\circ}=\frac{110^{\circ}}{2}=55^{\circ}$
Hence, the value of $x$ is $55^{\circ}$.
(ix)
Image
Let given triangle be $\triangle A B C$. Then, we have $A B=B C$ So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle B A C=\angle B C A=x$
From the figure,
$\angle B C A=\angle D C E=30^{\circ} \text { [vertically opposite angles] }$
From Eq. (i), $\angle B A C=\angle B C A$
$\Rightarrow \quad x=30^{\circ}$
Hence, the value of $x$ is $30^{\circ}$.

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