Question

In $\Delta ABC$, with usual notations $\frac{b\sin B - c\sin C}{\sin(B-C)} =$

• A. $b$
• B. $c$
• C. $a$
• D. $a + b + c$

Hint:

For abstract triangle identities, try testing a specific simple triangle like an equilateral triangle ($A=B=C=60^\circ$). However, here that leads to a $\frac{0}{0}$ form, so pick a $30^\circ-60^\circ-90^\circ$ right triangle to compute the values rapidly and match option choices!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a trigonometric fraction defined for a triangle $ABC$. We need to simplify the expression $\frac{b\sin B - c\sin C}{\sin(B-C)}$ using standard triangle laws and properties.

Step 2: Key Formula or Approach:
According to the Sine Rule for any triangle $ABC$: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$$ This allows us to substitute side lengths with their corresponding sine components: $a = k\sin A$, $b = k\sin B$, and $c = k\sin C$. We will also use the difference of squares identity $\sin^2 B - \sin^2 C = \sin(B+C)\sin(B-C)$.

Step 3: Detailed Explanation:
Let's substitute $b = k\sin B$ and $c = k\sin C$ into the numerator of our expression: $$\text{Numerator} = (k\sin B)\sin B - (k\sin C)\sin C = k\sin^2 B - k\sin^2 C = k(\sin^2 B - \sin^2 C)$$ Using the identity $\sin^2 B - \sin^2 C = \sin(B+C)\sin(B-C)$: $$\text{Numerator} = k\sin(B+C)\sin(B-C)$$ Now, substitute this simplified numerator back into the complete fraction: $$\text{Expression} = \frac{k\sin(B+C)\sin(B-C)}{\sin(B-C)}$$ Cancel the matching term $\sin(B-C)$ from both the top and bottom: $$\text{Expression} = k\sin(B+C)$$ In any triangle $ABC$, $A + B + C = \pi \implies B + C = \pi - A$. Substituting this gives: $$\text{Expression} = k\sin(\pi - A)$$ Since $\sin(\pi - A) = \sin A$: $$\text{Expression} = k\sin A$$ By referencing our initial Sine Rule definition, $k\sin A$ is exactly equal to the side length $a$.

Step 4: Final Answer:
The simplified value of the fraction is $a$, which corresponds to option (C).