Q. $ABC$ 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 $AD = 35$ cm, then $CD\$equal to
A) $35\sqrt{2}$ cm
B) $70\sqrt{2}$cm
C) $\frac{35\sqrt{3}}{2}$ cm
D) $35\sqrt{6}$ cm
Explanation
(a)
In $\mathrm{\Delta}\ BCD,\tan{15{^\circ} = \frac{BC}{BD}}$
$\Rightarrow \ \ \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{x}{x + 35}$
$\Rightarrow \frac{\sqrt{3 - 1}}{\sqrt{3} + 1} = \frac{x}{x + 35}$
$\Rightarrow \ \ \sqrt{3}x + 35\sqrt{3} - x - 35 = \sqrt{3}x + x$
$\Rightarrow \ \ x = \frac{35(\sqrt{3} - 1)}{2}$
$\therefore\ \ CD = \sqrt{\left( \frac{35}{2} \right)^{2}\{\left( \sqrt{3} + 1 \right)^{2} + \left( \sqrt{3} - 1 \right)^{2}\}}$
$= \frac{35}{2} \times 2\sqrt{2} = 35\sqrt{2}$ cm
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.