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Question Number 46283 by somil last updated on 23/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

$$PQ=12QR=5PR=\sqrt{{12}^2+5^2}=13$$$$$$$$$$$$\mathrm{\angle }DQP=\theta \mathrm{\angle }DQR=90-\theta$$$$\mathrm{\angle }DRQ=\theta \mathrm{\angle }QPD=90-\theta$$$$cot\mathrm{\angle }DQP$$$$=cot\theta$$$$=\frac{5}{12}$$$$cos\mathrm{\angle }DQR$$$$cos(90-\theta )$$$$=\frac{12}{13}plscheck...$$

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