Question
$ABCD$ is a trapezium in which $AB \| DC$ and $\angle\text{A}=\angle\text{B}=45^\circ.$ Find angles $C$ and $D$ of the trapezium.
✓
Answer
Given, $A B C D$ is a trapezium and whose parallel sides in the figure are $A B$ and $D C$. Since, $A B \| C D$ and $B C$ is transversal, then sum of two cointerior angles is $180^{\circ}$
.
$\therefore\ \angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{C}=180^\circ-\angle\text{B}=180^\circ-45^\circ$ [$\because\angle\text{B}=45^\circ$ given] $\Rightarrow\ \angle\text{C}=135^\circ$ Similarly, $\angle\text{A}+\angle\text{D}=180^\circ$ [sum of cointerior is 180^\circ ] $\Rightarrow\ \angle\text{D}=180^\circ-45^\circ$ [$\because\angle\text{A}=45^\circ$ given] $\Rightarrow\angle\text{D}=135^\circ$ Hence, angles $C$ and $D$ are $135^\circ$ each
.
$\therefore\ \angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{C}=180^\circ-\angle\text{B}=180^\circ-45^\circ$ [$\because\angle\text{B}=45^\circ$ given] $\Rightarrow\ \angle\text{C}=135^\circ$ Similarly, $\angle\text{A}+\angle\text{D}=180^\circ$ [sum of cointerior is 180^\circ ] $\Rightarrow\ \angle\text{D}=180^\circ-45^\circ$ [$\because\angle\text{A}=45^\circ$ given] $\Rightarrow\angle\text{D}=135^\circ$ Hence, angles $C$ and $D$ are $135^\circ$ each
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