Question

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

$\angle BPC$ (in degrees)

• A. ${65}^o$
• B. ${70}^o$
• C. ${75}^o$
• D. ${85}^o$

Answer 1

Answer: (C)

${75}^o$

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

Now in $△$ PCB,

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

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

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

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

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

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

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