Question

The equilateral triangle $ABP$ lies inside the square $ABCD$. Find $\angle CPD$.

• A. ${150}^o$
• B. ${145}^o$
• C. ${155}^o$
• D. ${160}^o$

Answer 1

The correct option is A ${150}^o$

Refer Figure

Given:

$ABCD$ is a square

$ABP$ is an equilateral triangle

Solution:

$AB=BP=BC$ ($△ABP$ is equilateral and $AB$ and $BC$ belong to a square, hence all are equal)

$\angle ABC=\angle ABP+\angle PBC=60°+\angle PBC$

$90°-60°=\angle PBC$

$\angle PBC=30°$

In $△BPC$

$BP=BC$

$\angle BPC=\angle BCP=\frac{180-\angle PBC}{2}=\frac{150°}{2}=75°$ ( angles opposite to equal sides of a triangle are equal)

Similarly, $\angle APD=75°$

$\angle APB+\angle BPC+\angle CPD+\angle APD=(60+75+75)°+\angle CPD=360°$

Hence, $\angle CPD=360°-210°=150°$