Question Number 172915 by mnjuly1970 last updated on 03/Jul/22
Commented by infinityaction last updated on 03/Jul/22
$$\mathrm{6}? \\ $$
Answered by mr W last updated on 03/Jul/22
Commented by mr W last updated on 03/Jul/22
$$\mathrm{cos}\:\gamma=\frac{\mathrm{4}−\mathrm{1}}{\mathrm{4}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${BD}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}×\mathrm{4}×\mathrm{5}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{65} \\ $$$${EB}=\sqrt{{BD}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\mathrm{8} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\mathrm{tan}\:{B}=\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\frac{\mathrm{4}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}}{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{8}}}=\frac{\mathrm{39}}{\mathrm{52}} \\ $$$${AC}={AB}×\mathrm{tan}\:{B}=\mathrm{8}×\frac{\mathrm{39}}{\mathrm{52}}=\mathrm{6}\:\checkmark \\ $$
Commented by Tawa11 last updated on 03/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by som(math1967) last updated on 04/Jul/22
Commented by som(math1967) last updated on 04/Jul/22
![let AC=x ∠DCO=∠OCB=θ ∴∠ACB=2θ AE=4−1=3 ,OE=(√(5^2 −3^2 ))=4 ∴CD=x−4 tanθ=(1/(x−4)) ,tan2θ=(8/x) tan2θ=((2tanθ)/(1−tan^2 θ)) (8/x)=((2/(x−4))/(((x−4)^2 −1)/((x−4)^2 ))) (8/x)=((2(x−4))/(x^2 −8x+15)) 4x^2 −32x+60=x^2 −4x 3x^2 −28x+60=0 3x^2 −18x−10x+60=0 (x−6)(3x−10)=0 (x−6)=0 [ ∵x>4 ∴x>((10)/3), (3x−10)≠0] ∴x=6](https://www.tinkutara.com/question/Q172964.png)
$${let}\:{AC}={x} \\ $$$$\angle{DCO}=\angle{OCB}=\theta \\ $$$$\therefore\angle{ACB}=\mathrm{2}\theta \\ $$$${AE}=\mathrm{4}−\mathrm{1}=\mathrm{3}\:\:,{OE}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }=\mathrm{4} \\ $$$$\therefore{CD}={x}−\mathrm{4} \\ $$$${tan}\theta=\frac{\mathrm{1}}{{x}−\mathrm{4}}\:\:,{tan}\mathrm{2}\theta=\frac{\mathrm{8}}{{x}} \\ $$$$\:{tan}\mathrm{2}\theta=\frac{\mathrm{2}{tan}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \theta} \\ $$$$\:\frac{\mathrm{8}}{{x}}=\frac{\frac{\mathrm{2}}{{x}−\mathrm{4}}}{\frac{\left({x}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{8}}{{x}}=\frac{\mathrm{2}\left({x}−\mathrm{4}\right)}{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{60}={x}^{\mathrm{2}} −\mathrm{4}{x} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{60}=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{18}{x}−\mathrm{10}{x}+\mathrm{60}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left(\mathrm{3}{x}−\mathrm{10}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)=\mathrm{0}\:\:\:\left[\:\because{x}>\mathrm{4}\:\:\therefore{x}>\frac{\mathrm{10}}{\mathrm{3}},\:\left(\mathrm{3}{x}−\mathrm{10}\right)\neq\mathrm{0}\right] \\ $$$$\therefore{x}=\mathrm{6} \\ $$
Commented by Tawa11 last updated on 04/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by infinityaction last updated on 04/Jul/22
Commented by infinityaction last updated on 04/Jul/22
$${DB}\:=\:\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \:}\:\:\:\:=\:\:\:\sqrt{\mathrm{65}} \\ $$$$\bigtriangleup{DEB} \\ $$$$\:\:{BE}\:\:=\:\:\sqrt{\left(\sqrt{\mathrm{65}}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }\:\:=\:\:\:\mathrm{8} \\ $$$$\bigtriangleup{ABC} \\ $$$$\:\:\left({x}+\mathrm{4}\right)^{\mathrm{2}} \:+\:\mathrm{8}^{\mathrm{2}} \:\:\:=\:\:\left({x}+\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\:\cancel{\:\:{x}^{\mathrm{2}} }+\mathrm{16}+\mathrm{8}{x}+\cancel{\mathrm{64}}\:=\cancel{\:{x}^{\mathrm{2}} }+\cancel{\mathrm{64}}+\mathrm{16}{x} \\ $$$$\:\:\:\mathrm{8}{x}\:=\:\:\mathrm{16}\:\:\Rightarrow\:\:{x}\:=\:\mathrm{2} \\ $$$${AC}\:=\:\mathrm{4}+{x}\:=\:\mathrm{4}+\mathrm{2}\:=\:\mathrm{6} \\ $$
Commented by Tawa11 last updated on 04/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$