Properties of Triangles 1 Question 8
8. In a $\triangle A B C, 2 a c \sin \frac{1}{2}(A-B+C)$ is equal to
(a) $a^{2}+b^{2}-c^{2}$
(b) $c^{2}+a^{2}-b^{2}$
(c) $b^{2}-c^{2}-a^{2}$
(d) $c^{2}-a^{2}-b^{2}$
$(2000,2 M)$
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Answer:
Correct Answer: 8. (a)
Solution:
- We know that, $A+B+C=180^{\circ}$
$\Rightarrow$
$$ A+C-B=180-2 B $$
Now, $2 a c \sin \frac{1}{2}(A-B+C)=2 a c \sin \left(90^{\circ}-B\right)$
$$ \begin{aligned} & =2 a c \cos B=\frac{2 a c \cdot\left(a^{2}+c^{2}-b^{2}\right)}{2 a c} \quad \text { [by cosine rule] } \\ & =a^{2}+c^{2}-b^{2} \end{aligned} $$
