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Find the values of the unknowns $x$ and $y$ in the following diagrams:
"
Given: Triangles in the above-given diagram with unknown angles $x$ and $y$.
To do: To find the values of unknown angles $x$ and $y$ in each case.
Solution:
For convenience, we name all the triangles given in the diagram as $\triangle ABC$.
$(i).\ \angle y+\angle ACD=180^{\circ}$ [linear pair]
$\Rightarrow \angle y\ +\ 120^{\circ}=180^{\circ}$
$\Rightarrow \angle y=60^{\circ}$
By using the angle sum property
In triangle $ABC$
$\angle A+\angle B+\ \angle C=180^{\circ}$ [$\angle A=x,\ \angle C=y$]
$\Rightarrow x+50^{\circ}+60^{\circ}=180^{\circ}$
$\Rightarrow x=180^{\circ}-110^{\circ}$
$\Rightarrow x=70^{\circ}$
$(ii).\ \angle EAD=\angle y$ [vertically opposite angles are the same]
$80^{\circ}=\angle y$
By using the angle sum property
In triangle $ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow 80^{\circ}+\ 50^{\circ}\ +x=180^{\circ}$
$\Rightarrow x=180^{\circ}-130^{\circ}$
$\Rightarrow x=50^{\circ}$
$(iii)$ By using the angle sum property
In triangle $ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow 50^{\circ}+60^{\circ}+y=180^{\circ}$
$\Rightarrow y=180^{\circ}-110^{\circ}$
$\Rightarrow y=70^{\circ}$
We know that, $x+y=180^{\circ}$ [linear pair]
$\Rightarrow x+\ 70^{\circ}=180^{\circ}$
$\Rightarrow x=110^{\circ}$
$(iv).\ x=\angle DCE$ [vertically opposite angles are the same]
$60^{\circ}=x$
By using the angle sum property
In trianfle $ABC$,
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow 30^{\circ}+60^{\circ}+y=180^{\circ}$
$\Rightarrow y=180^{\circ}-90^{\circ}$
$\Rightarrow y=90^{\circ}$
$(v).\ \angle EAD=y=90^{\circ}$ [vertically opposite angles are the same]
In triangle $ABC$,
By using the angle sum property,
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow x+x+90^{\circ}=180^{\circ}$
$\Rightarrow 2x=180^{\circ}-90^{\circ}$
$\Rightarrow 2x=90^{\circ}$
$\Rightarrow x=\frac{90^{\circ}}{2}=45^{\circ}$
$(vi)$. by using the angle sum property,
$y=x=\angle A=\angle B=\angle C$ [vertically opposite angles are the same]
In triangle $ABC$
$\Rightarrow x+x+x=180^{\circ}$
$\Rightarrow 3x=180^{\circ}$
$\Rightarrow x=\frac{180^{\circ}}{3}$
$\Rightarrow x=60^{\circ}$
Updated on: 10-Oct-2022
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