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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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