Question

In ΔABC, prove that $\frac{a\ sin(B - C)}{b^{2} - c^{2}}\ =\ \frac{b\ sin(C - A)}{c^{2} - a^{2}}\ =\ \frac{c\ sin(A - B)}{a^{2} - b^{2}}$.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

In triangle ABC,

1.A+B+C=180

2.(a/SinA)=(b/sinB)=(c/sinC)=2R

$3.{sin}^2A-{sin}^2B=sin(A+B).sin(A-B)$

$\frac{a.sin(B-C)}{{b}^2-{c}2}$

$=\frac{a.sin(B-C)}{{(2R.sinB)}^2-{(2R.sinC)}^2}$

$=\frac{a.sin(B-C)}{{4R}^2(sin^2B-{sin^2C})}$

$=\frac{2R.sinA.sin(B-C)}{{4R}^2(sin^2B-sin^2C)}$

$=\frac{sinA.sin(B-C)}{2R(sin^2B-sin^2C)}$

$=\frac{sinA.sin(B-C)}{2R.sin(B+C).sin(B-C)}$

$=\frac{sinA}{2R.sin(180-A)}$

$=\frac{sinA}{2R.sinA}\\\\=\frac{1}{2R}$

$\\\frac{a.sin(B-C)}{{b}^2-{c}^2}=\frac{1}{2R}$

similarly we can prove

$\frac{b.sin(C-A)}{{c}^2-{a}2}=\frac{1}{2R}$

$\frac{c.sin(A-B)}{{a}^2-{b}^2}=\frac{1}{2R}$$</p> <p>Hence\\\frac{a.sin(B-C)}{{b}^2-{c}2}=\frac{b.sin(C-A)}{{c}^2-{a}2}=\frac{c.sin(A-B)}{{a}^2-{b}2}$