SSC CGL Tier II 12 Apr 2015 Paper

(3) sin 15° = sin (45° – 30°)= sin 45° × cos 30° – cos 45° × sin 30°
$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$
$=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$
and cos 15° = cos (45°– 30°) = cos 45°. cos 30° + sin 45°. sin 30°
$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$
$=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$
∴ AB = AC sin 15° $=\frac{\sqrt{3}-1}{2\sqrt{2}}$ metre
BC = AC cos 15° $=\frac{\sqrt{3}+1}{2\sqrt{2}}$ metre
∴ Area of ΔABC $=\frac{1}{2}$ × AB × BC
$=\left(\frac{1}{2}\times \frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{3}+1}{2\sqrt{2}}\right)$ square metre
$=\left(\frac{3-1}{16}\right)$ square metre
$=\frac{1}{8}$ square metre.
$=\frac{10000}{8}=1250$ square $\mathrm{cm}.$