Question

In a triangle ABC with usual notations if \( b \sin C (b \cos C + c \cos B) = 42 \), then area of triangle ABC =}

• A. 42 sq. units
• B. 21 sq. units
• C. 24 sq. units
• D. 12 sq. units

Hint:

Projection rule: $a = b \cos C + c \cos B$. Area $= \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Use properties of triangles (projection rule) to simplify the given equation.
Step 2: Key Formula or Approach:
1. Projection Rule: \( a = b \cos C + c \cos B \).
2. Area of triangle \( \Delta = \frac{1}{2} ab \sin C \).
Step 3: Detailed Explanation:
The given equation is \( b \sin C (b \cos C + c \cos B) = 42 \).
By projection rule, \( (b \cos C + c \cos B) = a \).
So, \( b \sin C (a) = 42 \implies ab \sin C = 42 \).
We know Area \( \Delta = \frac{1}{2} ab \sin C \).
\[ \Delta = \frac{1}{2} (42) = 21 \text{ sq. units} \] Step 4: Final Answer:
The area is 21 sq. units.