Question Number 211190 by mr W last updated on 30/Aug/24
Answered by A5T last updated on 31/Aug/24
$${Let}\:\angle{BCD}=\angle{CBD}=\theta\Rightarrow\angle{DBA}=\mathrm{90}°−\theta \\ $$$$\Rightarrow\angle{ADB}=\mathrm{90}°−\theta\Rightarrow\angle{DAB}=\mathrm{2}\theta \\ $$$$\angle{BCA}=\angle{CAB}=\mathrm{45}°\Rightarrow\angle{ACD}=\mathrm{45}°−\theta\Rightarrow?=\mathrm{45}°−\mathrm{2}\theta \\ $$$${Let}\:{CD}={BD}={x}\:{and}\:{AD}={AB}={BC}={y} \\ $$$${Then}\:{AC}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xycos}\left(\mathrm{3}\theta+\mathrm{90}°\right)…\left({i}\right) \\ $$$${BD}^{\mathrm{2}} ={x}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xycos}\left(\theta\right)\Rightarrow{y}=\mathrm{2}{xcos}\theta…\left({ii}\right) \\ $$$$\left({ii}\right)\:{in}\:\left({i}\right)\Rightarrow\mathrm{4}{cos}^{\mathrm{2}} \theta=\mathrm{1}−\mathrm{4}\left({cos}\theta\right){cos}\left(\mathrm{3}\theta+\mathrm{90}\right) \\ $$$$\Rightarrow\theta=\mathrm{15}°\Rightarrow?=\mathrm{45}°−\mathrm{2}\theta=\mathrm{15}° \\ $$
Commented by mr W last updated on 31/Aug/24
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Answered by mm1342 last updated on 31/Aug/24
$$<{DBC}=\theta\:\Rightarrow<{ADB}=\mathrm{90}−\theta\:\:\&\:\:\:<{CDB}=\mathrm{180}−\mathrm{2}\theta\:\:\&\:\:<{DAB}=\mathrm{2}\theta \\ $$$$\frac{{CD}}{{BC}}=\frac{{sin}\theta}{{sin}\mathrm{2}\theta}\:=\frac{\mathrm{1}}{\mathrm{2}{cos}\theta}\:\:\:\&\:\:\:\frac{{BD}}{{AD}}=\frac{{sin}\mathrm{2}\theta}{{cos}\theta}=\mathrm{2}{sin}\theta \\ $$$$\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\theta=\mathrm{15}\:\:\checkmark \\ $$$$ \\ $$
Commented by mr W last updated on 31/Aug/24
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Answered by mr W last updated on 31/Aug/24
