Step 1: Understanding the Concept:
We need to verify that the three fundamental Pythagorean trigonometric identities hold true for the specific angle \(\theta = 30^\circ\). This involves substituting the known trigonometric values for 30\(^\circ\) into the equations and showing that the Left Hand Side (LHS) equals the Right Hand Side (RHS).
Step 2: Key Formula or Approach:
We need the standard trigonometric values for \(\theta = 30^\circ\):
\[\begin{array}{rl} \bullet & \text{\(\sin 30^\circ = \frac{1}{2}\)} \\ \bullet & \text{\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)} \\ \bullet & \text{\(\tan 30^\circ = \frac{1}{\sqrt{3}}\)} \\ \bullet & \text{\(\csc 30^\circ = 2\)} \\ \bullet & \text{\(\sec 30^\circ = \frac{2}{\sqrt{3}}\)} \\ \bullet & \text{\(\cot 30^\circ = \sqrt{3}\)} \\ \end{array}\]
Step 3: Detailed Explanation:
(i) Verify \(\sin^2\theta + \cos^2\theta = 1\)
LHS = \(\sin^2(30^\circ) + \cos^2(30^\circ)\)
\[ = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ = \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1 \]
RHS = 1.
Since LHS = RHS, the identity is verified.
(ii) Verify \(1 + \tan^2\theta = \sec^2\theta\)
LHS = \(1 + \tan^2(30^\circ)\)
\[ = 1 + \left(\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3} \]
RHS = \(\sec^2(30^\circ)\)
\[ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]
Since LHS = RHS, the identity is verified.
(iii) Verify \(1 + \cot^2\theta = \csc^2\theta\)
LHS = \(1 + \cot^2(30^\circ)\)
\[ = 1 + (\sqrt{3})^2 = 1 + 3 = 4 \]
RHS = \(\csc^2(30^\circ)\)
\[ = (2)^2 = 4 \]
Since LHS = RHS, the identity is verified.
Step 4: Final Answer:
All three trigonometric identities are successfully verified for \(\theta = 30^\circ\).
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.
In the figure given below, find RS and PS using the information given in \(\triangle\)PSR. 