# Solve this ?

Oct 17, 2017

Option $\left(B\right) \text{ " : " } {45}^{\circ}$

#### Explanation:

Solution 1 :

Let $A P = A Q = a , \implies P B = Q D = 1 - a , \mathmr{and} P Q = a \sqrt{2}$,
given $A P + A Q + P Q = 2$,
$\implies a + a + a \sqrt{2} = 2$,
$\implies a = 2 - \sqrt{2}$
$\implies P B = Q D = 1 - a = \sqrt{2} - 1$
$\implies y = {\tan}^{-} 1 \left(\sqrt{2} - 1\right) = {22.5}^{\circ}$
$\implies \angle P C Q = x = 90 - 2 y = 90 - 2 \cdot 22.5 = {45}^{\circ}$

Solution 2 :

$A P + A Q + P Q = 2$,
As $A Q + Q D = 1 , \mathmr{and} A P + P B = 1$,
$\implies A P + A Q + Q D + P B = 2$,
$\implies \textcolor{red}{P Q = P B + D Q}$
Now, extend $A B$ to $E$ such that $B E = D Q$,
$\implies \Delta C B E \mathmr{and} \Delta C D Q$ are congruent,
$\implies C E = C Q$
As angleDCB=90^@, => color(red)(angleQCE=90^@
$\textcolor{red}{P E} = P B + B E = \textcolor{red}{P B + D Q = P Q}$,
Now, as $C Q = C E , P Q = P E , \mathmr{and} C P$ is the common side,
$\implies \Delta C P Q \mathmr{and} \Delta C P E$ are congruent,
$\implies \angle P C Q = \angle P C E = \frac{\angle Q C E}{2} = \frac{90}{2} = {45}^{\circ}$

Oct 19, 2017

Let $\angle Q C D = \alpha$

$\angle P C B = \beta$

So $\angle P C Q = {90}^{\circ} - \left(\alpha + \beta\right)$

Now for $\Delta Q C D , \angle Q D C = {90}^{\circ} \mathmr{and} D C = 1 u n i t$
So $D Q = D C \tan \alpha = \tan \alpha$

And for $\Delta P C B , \angle P B C = {90}^{\circ} \mathmr{and} B C = 1 u n i t$
So $P B = B C \tan \beta = \tan \beta$

Hence for $\Delta A P Q$

$A Q = 1 - \tan \alpha$

$A P = 1 - \tan \beta$

and

$P Q = \sqrt{{\left(1 - \tan \alpha\right)}^{2} + {\left(1 - \tan \beta\right)}^{2}}$

Given

$A Q + A P + P Q = 2$

$\implies 1 - \tan \alpha + 1 - \tan \beta + \sqrt{{\left(1 - \tan \alpha\right)}^{2} + {\left(1 - \tan \beta\right)}^{2}} = 2$

$\implies \sqrt{{\left(1 - \tan \alpha\right)}^{2} + {\left(1 - \tan \beta\right)}^{2}} = \tan \alpha + \tan \beta$

$\implies 1 - 2 \tan \alpha + {\tan}^{2} \alpha + 1 - 2 \tan \beta + {\tan}^{2} \beta = {\tan}^{2} \alpha + {\tan}^{2} \beta + 2 \tan \alpha \tan \beta$

$\implies 1 - \tan \alpha - \tan \beta = \tan \alpha \tan \beta$

$\implies \tan \alpha + \tan \beta = 1 - \tan \alpha \tan \beta$

$\implies \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 1 = \tan {45}^{\circ}$

$\implies \tan \left(\alpha + \beta\right) = \tan {45}^{\circ}$

$\implies \alpha + \beta = {45}^{\circ}$

Hence $\angle P C Q = {90}^{\circ} - \left(\alpha + \beta\right) = {90}^{\circ} - {45}^{\circ} = {45}^{\circ}$