- A$63^\circ$
- ✓$117^\circ$
- C$17^\circ$
- D$153^\circ$
$\therefore\ \angle\text{BOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow\ 63^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\ \angle\text{BOC}=117^\circ$
66 questions · 50 auto-graded MCQ + 16 self-marked written.






Let angle supplement of $3 x^{\circ}$ be $Z^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow \angle \mathrm{AHF}+\angle \mathrm{FHB}=180^{\circ}$
$\Rightarrow z^{\circ}+3 x^{\circ}=180^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
Now,
$x^{\circ}+y^{\circ}=180^{\circ}$
Also,
$x^{\circ}=z^{\circ} \text { [Correspondence angles] }$
$\Rightarrow x^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow 4 x^{\circ}=180^{\circ}$
$\Rightarrow x^{\circ}=45^{\circ}$
$x^{\circ}+y^{\circ}=180^{\circ}$
$\Rightarrow y^{\circ}=180^{\circ}-x^{\circ}=180^{\circ}-45^{\circ}=135^{\circ}$.


From figure,
$\angle\text{EPR}=\angle\text{PQS}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{PQS}=\text{x}^\circ$
Also,
$\angle\text{PQS}=\angle\text{RSD}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{RSD}=\text{x}^\circ$
Now,
$\angle\text{RSD}+\text{y}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ+2\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{y}^\circ=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\ \text{y}^\circ=90^\circ-\frac{\text{x}^\circ}{2}$






































