Question
In the adjoining figure, three coplanar lines $AB, CD$ and $EF$ intersect at a point $O$, forming angles as shown. Find the values of $x, y, z$ and $t$. 
✓
Answer
We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore\angle\text{BOD}=\angle\text{AOC}=90^\circ$
Hence, $t = 90^\circ$ Also, $\angle\text{DOF}=\angle\text{COE}=50^\circ$
Hence, $z = 50^\circ$ Since, $AOB$ is a straight line,
we have: $\angle\text{AOC}+\angle\text{COE}+\angle\text{BOE}=180^\circ$
$\Rightarrow 90 + 50 + y = 180^\circ $
$\Rightarrow 140 + y = 180^\circ $
$\Rightarrow y = 40^\circ $
Also, $\angle\text{BOE}=\angle\text{AOF}=40^\circ$
Hence, $x = 40^\circ $ $\therefore$ $x = 40^\circ , y = 40^\circ , z = 50^\circ $ and $t = 90^\circ$
$\therefore\angle\text{BOD}=\angle\text{AOC}=90^\circ$
Hence, $t = 90^\circ$ Also, $\angle\text{DOF}=\angle\text{COE}=50^\circ$
Hence, $z = 50^\circ$ Since, $AOB$ is a straight line,
we have: $\angle\text{AOC}+\angle\text{COE}+\angle\text{BOE}=180^\circ$
$\Rightarrow 90 + 50 + y = 180^\circ $
$\Rightarrow 140 + y = 180^\circ $
$\Rightarrow y = 40^\circ $
Also, $\angle\text{BOE}=\angle\text{AOF}=40^\circ$
Hence, $x = 40^\circ $ $\therefore$ $x = 40^\circ , y = 40^\circ , z = 50^\circ $ and $t = 90^\circ$
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