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Question Number 28762 by naka3546 last updated on 30/Jan/18

Commented by naka3546 last updated on 30/Jan/18

$$PF?$$

Commented by naka3546 last updated on 30/Jan/18

$$PF?$$

Commented by ajfour last updated on 30/Jan/18

Commented by ajfour last updated on 30/Jan/18

$$\tan \alpha =2$$$$\mathit{x}\tan \mathit{\alpha }=6$$$$\Rightarrow \mathit{x}=\frac{6}{\tan \mathit{\alpha }}=\frac{6}{2}=3.$$

Commented by mrW2 last updated on 30/Jan/18

$${That}^{\prime }snice!$$

Commented by ajfour last updated on 30/Jan/18

$$thanksSir!$$

Answered by mrW2 last updated on 30/Jan/18

Commented by mrW2 last updated on 30/Jan/18

$$\frac{BP}{\sin (45-\theta )}=\frac{AB}{\sin (45+45-\theta )}=\frac{AP}{\sin 45}$$$$\Rightarrow BP=\frac{\sin (45-\theta )}{\sin 45}\times 6=(\cos \theta -\sin \theta )\times 6$$$$\Rightarrow AB=\frac{\sin (90-\theta )}{\sin 45}\times 6=\sqrt{2}\cos \theta \times 6$$$$$$$$\frac{DE}{\sin \theta }=\frac{PD}{\sin (45+\theta )}$$$$DE=\frac{AB}{2}$$$$\Rightarrow PD=\frac{\sin (45+\theta )}{\sin \theta }\times \frac{\sqrt{2}\cos \theta \times 6}{2}=\frac{(\sin \theta +\cos \theta )\cos \theta }{\sin \theta }\times \frac{6}{2}$$$$$$$$BP+PD=\sqrt{2}AB$$$$\Rightarrow (\cos \theta -\sin \theta )\times 6+\frac{(\sin \theta +\cos \theta )\cos \theta }{\sin \theta }\times \frac{6}{2}=\sqrt{2}\times \sqrt{2}\cos \theta \times 6$$$$\Rightarrow 2(\cos \theta -\sin \theta )+\frac{(\sin \theta +\cos \theta )\cos \theta }{\sin \theta }=4\cos \theta$$$$\Rightarrow \frac{(\sin \theta +\cos \theta )\cos \theta }{\sin \theta }=2(\sin \theta +\cos \theta )$$$$\Rightarrow \frac{\cos \theta }{\sin \theta }=2$$$$\Rightarrow \tan \theta =\frac{1}{2}$$$$$$$$\Rightarrow PF=AP\tan \theta =6\times \frac{1}{2}=3$$

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