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Question Number 47831 by somil last updated on 15/Nov/18

Commented by somil last updated on 16/Nov/18

$$iknow$$

Answered by $@ty@m last updated on 15/Nov/18

$$\tan 45=\frac{PQ}{QR}$$$$\Rightarrow PQ=QR=x,say...\left(1\right)$$$$\tan 30=\frac{x}{x+30}$$$$\frac{1}{\sqrt{3}}=\frac{x}{x+30}$$$$x\sqrt{3}=x+30$$$$x=\frac{30}{\sqrt{3}-1}$$$$x=15(\sqrt{3}+1)=15\times 2.732$$$$x=40.98cm$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18

$$tan{45}^o=\frac{PQ}{QR}=1soPQ=QR$$$$tan{30}^o=\frac{PQ}{QD}=\frac{PQ}{QR+RD}=\frac{PQ}{PQ+RD}=\frac{1}{\sqrt{3}}$$$$PQ\sqrt{3}=PQ+RD$$$$PQ\sqrt{3}=PQ+30$$$$PQ\sqrt{3}-PQ=30$$$$PQ(\sqrt{3}-1)=30$$$$PQ=\frac{30}{\sqrt{3}-1}=\frac{30(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{30(\sqrt{3}+1)}{3-1}=15(\sqrt{3}+1)$$$$P\stackrel{}{Q}=QR=15(\sqrt{3}+1)=40.98cm$$

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