Question

The given figure shows a square $ABCD$ and an equilateral triangle $ABP$. Calculate

$\angle PCD$ (in degrees)

• A. ${20}^o$
• B. ${15}^o$
• C. ${25}^o$
• D. ${40}^o$

Answer 1

Answer: (B)

${15}^o$

Given that ABCD is a square & $△$ APB is an equilateral triangle.

Hence $\angle PAB=\angle BAO=\angle ABP=\angle BPA={60}^o$[ All interior angles of equilateral triangle are equal to ${60}^o$]

So $\angle PBC=\angle ABC-\angle ABP={90}^o-{60}^o={30}^o$

Since PB=AB=BC so $△$ PCB is a isosceles triangle.

So we get $\angle BCP=\angle BPC$[ Base angles of isosceles triangle are equal]

Now in $△$ PCB,

$\angle BPC+\angle BCP+\angle PBC={180}^o$

$\Rightarrow 2\angle BCP+\angle PBC={180}^o$

$\Rightarrow 2\angle BCP+{30}^o={180}^o$

$\Rightarrow 2\angle BCP={180}^o-{30}^o$

$\Rightarrow 2\angle BCP={150}^o$

$\Rightarrow \angle BCP=\frac{{150}^o}{2}$

$\Rightarrow \angle BCP={75}^o$

Now $\angle PCD=\angle BCD-\angle BCP$

$\Rightarrow \angle PCD={90}^o-{75}^o$

$\Rightarrow \angle PCD={15}^o$