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 \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60\(^\circ\). To find the height of tree complete the activity. (\(\sqrt{3} = 1.73\)) 
Activity :
In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
Angle of elevation (\(\theta\)) = \(\angle\)BCA = 60\(^\circ\)
tan \(\theta\) = \(\frac{\boxed{\phantom{AB}}}{BC}\) \(\dots\) (I)
tan 60\(^\circ\) = \(\boxed{\phantom{\sqrt{3}}}\) \(\dots\) (II)
\(\frac{AB}{BC} = \sqrt{3}\) \(\dots\) (From (I) and (II))
AB = BC \(\times\) \(\sqrt{3}\) = 10\(\sqrt{3}\)
AB = 10 \(\times\) 1.73 = \(\boxed{\phantom{17.3}}\)
\(\therefore\) height of the tree is \(\boxed{\phantom{17.3}}\) m.