Step 1: Understanding the Concept:
We need to use the fundamental Pythagorean trigonometric identity that relates the sine and cosine of an angle.
Step 2: Key Formula or Approach:
The Pythagorean identity is:
\[ \sin^2\theta + \cos^2\theta = 1 \]
We can rearrange this to solve for \(\cos\theta\):
\[ \cos^2\theta = 1 - \sin^2\theta \]
\[ \cos\theta = \sqrt{1 - \sin^2\theta} \]
Step 3: Detailed Explanation:
Given:
\[ \sin\theta = \frac{11}{61} \]
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ \left(\frac{11}{61}\right)^2 + \cos^2\theta = 1 \]
\[ \frac{11^2}{61^2} + \cos^2\theta = 1 \]
\[ \frac{121}{3721} + \cos^2\theta = 1 \]
Now, solve for \(\cos^2\theta\):
\[ \cos^2\theta = 1 - \frac{121}{3721} \]
\[ \cos^2\theta = \frac{3721 - 121}{3721} \]
\[ \cos^2\theta = \frac{3600}{3721} \]
Take the square root of both sides. Since it is not specified, we assume \(\theta\) is in the first quadrant where cosine is positive.
\[ \cos\theta = \sqrt{\frac{3600}{3721}} \]
\[ \cos\theta = \frac{\sqrt{3600}}{\sqrt{3721}} \]
\[ \cos\theta = \frac{60}{61} \]
Step 4: Final Answer:
The value of \(\cos\theta\) is \( \frac{60}{61} \).
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.