In the figure given below, find RS and PS using the information given in \(\triangle\)PSR. 
Step 1: Understanding the Concept:
This problem involves solving a right-angled triangle using trigonometric ratios or the properties of a 30-60-90 triangle. Given one side and one angle, we can find the lengths of the other two sides.
Step 2: Key Formula or Approach:
We can use two methods:
Method 1: Trigonometric Ratios
\[\begin{array}{rl} \bullet & \text{\( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)} \\ \bullet & \text{\( \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)} \\ \end{array}\]
Method 2: 30-60-90 Triangle Theorem
The sides of a 30-60-90 triangle are in the ratio \( 1 : \sqrt{3} : 2 \).
\[\begin{array}{rl} \bullet & \text{The side opposite 30\(^\circ\) is half the hypotenuse.} \\ \bullet & \text{The side opposite 60\(^\circ\) is \( \sqrt{3} \) times the side opposite 30\(^\circ\).} \\ \end{array}\]
Step 3: Detailed Explanation:
Given:
\[\begin{array}{rl} \bullet & \text{In \(\triangle\)PSR, \(\angle\)S = 90\(^\circ\).} \\ \bullet & \text{Hypotenuse PR = 12.} \\ \bullet & \text{\(\angle\)P = 30\(^\circ\).} \\ \bullet & \text{Therefore, \(\angle\)R = 180\(^\circ\) - 90\(^\circ\) - 30\(^\circ\) = 60\(^\circ\).} \\ \end{array}\]
Using Method 2 (30-60-90 Triangle Theorem):
1. Find RS: RS is the side opposite the 30\(^\circ\) angle (\(\angle\)P).
According to the theorem, the side opposite the 30\(^\circ\) angle is half the hypotenuse.
\[ RS = \frac{1}{2} \times PR = \frac{1}{2} \times 12 = 6 \]
2. Find PS: PS is the side opposite the 60\(^\circ\) angle (\(\angle\)R).
According to the theorem, the side opposite the 60\(^\circ\) angle is \( \sqrt{3} \) times the side opposite the 30\(^\circ\) angle.
\[ PS = \sqrt{3} \times RS = \sqrt{3} \times 6 = 6\sqrt{3} \]
Step 4: Final Answer:
The length of RS is 6 and the length of PS is 6\(\sqrt{3}\).
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.