In the given figure, point D divides the side BC of $\displaystyle \triangle \mathrm{ABC}$ in the ratio $\displaystyle 1$ : $\displaystyle 2$ . Find length AD.

Coordinates of point \(\displaystyle \mathrm{D}=\left(\frac{4-4}{3}, \frac{2+2}{3}\right)\) i.e. \(\displaystyle \left(0, \frac{4}{3}\right)\) \(\displaystyle \mathrm{AD}=\sqrt{(1-0)^{2}+\left(5-\frac{4}{3}\right)^{2}}=\frac{\sqrt{130}}{3}\) units Sol. \(\displaystyle \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}=\frac{\left(\dfrac{5}{2}\right)-\dfrac{1}{\sqrt{3}}}{\left(\dfrac{1}{\sqrt{2}}\right)^{2}}\) \(\displaystyle =\frac{1}{4 \sqrt{3}}\) or \(\displaystyle \frac{\sqrt{3}}{12}\) Sol. \(\displaystyle \tan (\mathrm{A}+2 \mathrm{~B})=\sqrt{3} \Rightarrow \mathrm{~A}+2 \mathrm{~B}=60^{\circ}\) \(\displaystyle \sin (2 A+B)=\frac{1}{\sqrt{2}} \Rightarrow 2 A+B=45^{\circ}\) On solving above equations, \(\displaystyle \mathrm{A}=10^{\circ}, \mathrm{B}=25^{\circ}\)