Question

$ABC$ is a triangle with $B=90$ and $A=30$. $P,Q$ and $R$ are on $AB,BC$ and $CA$ respectively such that $PQR$ is an equilateral triangle. If $Q$ is the midpoint of $BC$ and $BC=4$, then the side of the equilateral triangle $PQR$ is $\sqrt{K}.$ The value of $K$ is equal to

Answer 1

${\theta }_1+{\theta }_2={120}^0$
also ${\theta }_2+{\theta }_3={120}^0$
Hence ${\theta }_1-{\theta }_3=0$ ... (i)
${\theta }_1={\theta }_3-\theta$ (say)
now in $\Delta QBP,sin\theta =\frac{x}{y}\Rightarrow y=\frac{x}{sin{\theta }_1}$ ...... (ii)
also in $\Delta CQR,\frac{2}{sin{\theta }_3}=\frac{y}{sin{60}^0}$
$\Rightarrow \frac{2}{sin{\theta }_3}=\frac{2y}{\sqrt{3}}\Rightarrow y=\frac{\sqrt{3}}{sin{\theta }_3}$ ..... (iii)
From (i), (ii) and (iii), we get
therefore $\frac{x}{sin{\theta }_1}=\frac{\sqrt{3}}{sin{\theta }_3}\Rightarrow x=\sqrt{3}$
Now $x^2+4=y^2\Rightarrow 7=y^2\Rightarrow y=\sqrt{7}\Rightarrow K=7$