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: Projection Rule
$b \cos C + c \cos B = a$.
Step 2: Substitution
$b \sin C (a) = 42 \implies ab \sin C = 42$.
Step 3: Area Formula
Area $(\triangle ABC) = \frac{1}{2} ab \sin C$.
Area $= \frac{1}{2} (42) = 21$.
Step 4: Conclusion
The area is 21 sq. units.
Final Answer:(B)