In the given figure, EAD BCD. Ray FAC cuts ray EAD at a point A s… - Vidyadip

MCQ

In the given figure, $EAD \perp BCD$. Ray $FAC$ cuts ray $EAD$ at a point A such that $\angle EAF =30^{\circ}$. Also, in $\triangle BAC , \angle BAC = x ^{\circ}$ and $\angle ABC =( x +10)^{\circ}$. Then, the value of $x$ is:

A
$30$
B
$20$
C
$35$
✓
$25$

✓

Answer

Correct option: D.

$25$

In the given figure $\angle\text{CAD}=\angle\text{EAF}$ (Vertically opposite angels)$\therefore \angle\text{CAD}=30^\circ$
In $\triangle\text{ABD},$
$\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}=180^\circ$ (Angle sum property)
$\Rightarrow (\text{x}+10)^\circ+(\text{x}^\circ+30^\circ)+90^\circ=180^\circ$
$\Rightarrow 2\text{x}+130^\circ=180^\circ$
$\Rightarrow 2\text{x}=180^\circ−130^\circ=50^\circ$
$\Rightarrow \text{x} = 25$
Thus, the value of $x$ is $25.$

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