Properties Of Triangles Ques 23
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) $c^{2}-a^{2}-b^{2}$
$(2000,2 M)$
Show Answer
Answer:
Correct Answer: 23.(b)
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} $
BETA
