question_answer

ABC is a right angled isosceles triangle with\[\angle B=90{}^\circ \]. If D is a point on AB so that\[\angle CDB=15{}^\circ \]and, if\[AD=35\text{ }cm,\]then CD is equal to:

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 :

In \[\Delta BCD.\] \[\tan 15{}^\circ =\frac{BC}{BD}\] \[\Rightarrow \]               \[\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{x}{x+35}\] \[\Rightarrow \]               \[\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{x}{x+35}\] \[\Rightarrow \]               \[\sqrt{3}x+35\sqrt{3}-x-35=\sqrt{3}x+x\] \[\Rightarrow \]               \[2x=35(\sqrt{3}-1)\] \[\Rightarrow \]               \[x=\frac{35(\sqrt{3}-1)}{2}\] \[\therefore \]\[CD=\frac{35}{2}\sqrt{{{\left( \frac{\sqrt{3}+1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}-1}{2} \right)}^{2}}}\]                 \[=\frac{35}{2}\times 2\sqrt{2}=35\sqrt{2}\,cm\]