Question

In the following diagram , ABCD is a square and APB is an equilateral triangle
Find the angles of $$ \Delta DPC $$

Answer 1

To find angles of $$\Delta DPC $$
$$AP = PB = AB$$ [ $$\Delta APB$$ is an equilateral triangle ]
$$AB = BC = CD = DA$$ [Sides of square ABCD ]
So, $$AP = DA$$ and $$PB = BC$$
In$$ \Delta APD $$
$$AP = DA$$
$$ \angle ADP + \angle APD + 30^{o} = 180^{o}$$
$$\Rightarrow \angle ADP + \angle ADP = 180^{0} - 30^{o} $$

$$\Rightarrow 2\angle ADP=150^0$$

$$\Rightarrow \angle ADP =75^0$$

We have $$\angle PDC=\angle D-\angle ADP$$
$$\Rightarrow \angle PDC=90^0-75^0$$
$$\Rightarrow \angle PDC=15^0$$
In $$\triangle BPC$$
$$PB=BC$$
$$\therefore \angle PCB=\angle BPC$$ ...[Angle opposite to equal sides are equal ]
$$\angle PCB+\angle BPC+\angle CBP=180^0$$ ...[Sum of Angles of a triangle]
$$\angle PCB+\angle PCB+30^0=180^0$$
$$2 \angle PCB=180^0-30^0$$
$$\angle PCB=75^0$$
We have $$\angle PCD=\angle C-\angle PCB$$
$$\Rightarrow \angle PCD=90^0-75^0$$
$$\Rightarrow \angle PCD=15^0$$
In $$\triangle DPC$$,
$$\angle PDC=15^0$$
$$\angle PCD=15^0$$
$$\angle PCD+\angle PDC+\angle DPC=180^0$$ ...[Sum of angles of a triangle]
$$\Rightarrow 15^0+15^0+\angle DPC=180^0$$
$$\angle DPC=180^0-30^0$$
$$\angle DPC=150^0$$
$$\therefore$$ Angles of $$\triangle DPC$$ are: $$15^0,150^0,15^0$$