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Question-46283




Question Number 46283 by somil last updated on 23/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
PQ=12  QR=5  PR=(√(12^2 +5^2 )) =13      ∠DQP=θ  ∠DQR=90−θ  ∠DRQ=θ   ∠QPD=90−θ  cot∠DQP  =cotθ  =(5/(12))  cos∠DQR  cos(90−θ)  =((12)/(13))   pls check...
$${PQ}=\mathrm{12}\:\:{QR}=\mathrm{5}\:\:{PR}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:=\mathrm{13} \\ $$$$ \\ $$$$ \\ $$$$\angle{DQP}=\theta\:\:\angle{DQR}=\mathrm{90}−\theta \\ $$$$\angle{DRQ}=\theta\:\:\:\angle{QPD}=\mathrm{90}−\theta \\ $$$${cot}\angle{DQP} \\ $$$$={cot}\theta \\ $$$$=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${cos}\angle{DQR} \\ $$$${cos}\left(\mathrm{90}−\theta\right) \\ $$$$=\frac{\mathrm{12}}{\mathrm{13}}\:\:\:{pls}\:{check}… \\ $$

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