Question

reflex$\angle APC$

• A. ${225}^{\circ }$
• B. ${405}^{\circ }$
• C. ${185}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is A ${225}^{\circ }$

$GiventhatABCDisasquare\&△APBisanequilateraltriangle.Hence\angle PAB=\angle BAO=\angle ABP=\angle BPA={60}^o\left[Allinterioranglesofequilateral△areequalto{60}^o\right]So\angle PBC=\angle ABC-\angle ABP={90}^o-{60}^o={30}^{o}SincePB=AB=BCso△PCBisaisoscelestriangle.Soweget\angle BPC=\angle BCP\left[Baseanglesofisoscelestriangleareequal\right]Nowin△PCB,\angle BPC+\angle BCP+\angle PBC={180}^o\Rightarrow 2\angle BPC+\angle PBC={180}^o\Rightarrow 2\angle BPC+{30}^o={180}^o\Rightarrow 2\angle BPC={180}^o-{30}^o\Rightarrow 2\angle BPC={150}^o\Rightarrow \angle BPC=\frac{{150}^o}{2}\Rightarrow \angle BPC={75}^{o}NowreflexAPC={360}^o-\angle BPA-\angle BPC\Rightarrow reflexAPC={360}^o-{60}^o-{75}^o\Rightarrow reflexAPC={360}^o-135\Rightarrow reflexAPC={225}^o$