question_answer

\[ABC\] is a right angled isosceles triangle with \[\angle B={{90}^{o}}\]. If D is a point on \[AB\] so that \[\angle DCB={{15}^{o}}\] and if \[AD=35cm\], then \[CD=\] [Kerala (Engg.) 2005]

A)  \[35\sqrt{2}\]cm

B) \[70\sqrt{2}cm\]

C) \[\frac{35\sqrt{3}}{2}cm\]

D) \[35\sqrt{6}\]cm

E) \[\frac{35\sqrt{2}}{2}cm\]

Correct Answer: A

Solution :

\[\angle DCB={{15}^{o}}\] \[\angle CAB={{45}^{o}}\] and \[\angle CDB={{75}^{o}}\] Let \[BD=x\]and \[AD=35\]cm. \[\tan \angle CAB=\frac{CB}{AB}\]Þ \[\tan {{45}^{o}}=\frac{CB}{35+x}\] \ \[\tan {{75}^{o}}=\frac{CB}{DB}=\frac{CB}{x}\]Þ \[CB=x\tan {{75}^{o}}\] \[CB=(35+x)\tan {{45}^{o}}\]= \[x\tan {{75}^{o}}\] Þ \[x=\frac{35\tan {{45}^{o}}}{\tan {{75}^{o}}-\tan {{45}^{o}}}\]= \[\frac{35}{\tan {{75}^{o}}-1}\] But \[\cos {{75}^{o}}=\frac{x}{CD}\] \[CD=\frac{x}{\cos {{75}^{o}}}\]\[=\frac{1}{\cos {{75}^{o}}}\times \frac{35}{\tan {{75}^{o}}-1}=\frac{35}{\sin {{75}^{o}}-\cos {{75}^{o}}}\]       \[=\frac{35}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}}=\frac{35}{\frac{2}{2\sqrt{2}}}\]\[=35\sqrt{2}\]cm.