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Question-47831




Question Number 47831 by somil last updated on 15/Nov/18
Commented by somil last updated on 16/Nov/18
i know
$${i}\:{know}\: \\ $$
Answered by $@ty@m last updated on 15/Nov/18
tan 45=((PQ)/(QR))  ⇒PQ=QR=x, say ...(1)  tan 30=(x/(x+30))  (1/( (√3)))=(x/(x+30))  x(√3)=x+30  x=((30)/( (√3)−1))  x=15((√3)+1)=15×2.732  x=40.98 cm
$$\mathrm{tan}\:\mathrm{45}=\frac{{PQ}}{{QR}} \\ $$$$\Rightarrow{PQ}={QR}={x},\:{say}\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\mathrm{30}=\frac{{x}}{{x}+\mathrm{30}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{{x}}{{x}+\mathrm{30}} \\ $$$${x}\sqrt{\mathrm{3}}={x}+\mathrm{30} \\ $$$${x}=\frac{\mathrm{30}}{\:\sqrt{\mathrm{3}}−\mathrm{1}} \\ $$$${x}=\mathrm{15}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)=\mathrm{15}×\mathrm{2}.\mathrm{732} \\ $$$${x}=\mathrm{40}.\mathrm{98}\:{cm} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
tan45^o =((PQ)/(QR))=1 so PQ=QR  tan30^o =((PQ)/(QD))=((PQ)/(QR+RD))=((PQ)/(PQ+RD))=(1/( (√3) ))  PQ(√3) =PQ+RD  PQ(√3) =PQ+30  PQ(√3) −PQ=30  PQ((√3) −1)=30  PQ=((30)/( (√3) −1))=((30((√3) +1))/(((√3) −1)((√3) +1)))=((30((√3) +1))/(3−1))=15((√3) +1)  PQ^ =QR=15((√3) +1)=40.98cm
$${tan}\mathrm{45}^{{o}} =\frac{{PQ}}{{QR}}=\mathrm{1}\:{so}\:{PQ}={QR} \\ $$$${tan}\mathrm{30}^{{o}} =\frac{{PQ}}{{QD}}=\frac{{PQ}}{{QR}+{RD}}=\frac{{PQ}}{{PQ}+{RD}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:} \\ $$$${PQ}\sqrt{\mathrm{3}}\:={PQ}+{RD} \\ $$$${PQ}\sqrt{\mathrm{3}}\:={PQ}+\mathrm{30} \\ $$$${PQ}\sqrt{\mathrm{3}}\:−{PQ}=\mathrm{30} \\ $$$${PQ}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)=\mathrm{30} \\ $$$${PQ}=\frac{\mathrm{30}}{\:\sqrt{\mathrm{3}}\:−\mathrm{1}}=\frac{\mathrm{30}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)}{\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)}=\frac{\mathrm{30}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)}{\mathrm{3}−\mathrm{1}}=\mathrm{15}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right) \\ $$$${P}\overset{} {{Q}}={QR}=\mathrm{15}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)=\mathrm{40}.\mathrm{98}{cm} \\ $$

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