Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 69272 by TawaTawa last updated on 22/Sep/19

Answered by $@ty@m123 last updated on 22/Sep/19

Let ∠DPC=θ ⇒∠Q=θ−30 Let CP=x In △DCP, tan θ=((√3)/x) ...(1) In △DCQ, tan (θ−30)=((√3)/(x+2)) ((tan θ−tan 30)/(1+tan θ.tan 30))=((√3)/(x+2)) ((((√3)/x)−(1/(√3)))/(1+(1/x)))=((√3)/(x+2)) (((3−x)/((√3) x))/((x+1)/x))=((√3)/(x+2)) ((3−x)/((√3)(x+1)))=((√3)/(x+2)) (3−x)(x+2)=3(x+1) 3x−x^2 +6−2x=3x+3 x^2 +2x−3=0 (x+3)(x−1)=0 x=−3, 1 x cannot be negative. ∴ x=1

$${Let}\:\angle{DPC}=\theta \\ $$$$\Rightarrow\angle{Q}=\theta−\mathrm{30} \\ $$$${Let}\:{CP}={x} \\ $$$${In}\:\bigtriangleup{DCP}, \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{\mathrm{3}}}{{x}}\:...\left(\mathrm{1}\right) \\ $$$${In}\:\bigtriangleup{DCQ}, \\ $$$$\mathrm{tan}\:\left(\theta−\mathrm{30}\right)=\frac{\sqrt{\mathrm{3}}}{{x}+\mathrm{2}} \\ $$$$\frac{\mathrm{tan}\:\theta−\mathrm{tan}\:\mathrm{30}}{\mathrm{1}+\mathrm{tan}\:\theta.\mathrm{tan}\:\mathrm{30}}=\frac{\sqrt{\mathrm{3}}}{{x}+\mathrm{2}} \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}}{{x}}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}=\frac{\sqrt{\mathrm{3}}}{{x}+\mathrm{2}} \\ $$$$\frac{\frac{\mathrm{3}−{x}}{\sqrt{\mathrm{3}}\:{x}}}{\frac{{x}+\mathrm{1}}{{x}}}=\frac{\sqrt{\mathrm{3}}}{{x}+\mathrm{2}} \\ $$$$\frac{\mathrm{3}−{x}}{\sqrt{\mathrm{3}}\left({x}+\mathrm{1}\right)}=\frac{\sqrt{\mathrm{3}}}{{x}+\mathrm{2}} \\ $$$$\left(\mathrm{3}−{x}\right)\left({x}+\mathrm{2}\right)=\mathrm{3}\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{3}{x}−{x}^{\mathrm{2}} +\mathrm{6}−\mathrm{2}{x}=\mathrm{3}{x}+\mathrm{3} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{3},\:\mathrm{1} \\ $$$${x}\:{cannot}\:{be}\:{negative}. \\ $$$$\therefore\:{x}=\mathrm{1} \\ $$

Commented by TawaTawa last updated on 22/Sep/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com