Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k:1.
Therefore, \(-1=\frac{6k-3}{k+1}\)
-k-1=6k-3
7k=2
k=\(\frac{2}{7}\)
Therefore,the required ratio is 2:7
What is the angle between the hour and minute hands at 4:30?
In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$. 
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 