\frac{\sin B}{\sin C}=\frac{c-a \cos B}{b-a \cos C}

1.	\frac{\sin B}{\sin C}=\frac{c-a \cos B}{b-a \cos C}
Answer» Solution: Given: ΔABC Prove that sinb/sinc = c- acosb/b-acosc Use cosine formulas: cosb = (a²+c² - b²)/2ac and cosc = (a² - c² + b² )/ 2ab RHS: c- acosb/b-acosc = (c- a (a²+c² - b²)/2ac)/ (b - a (a² - c² + b² )/ 2ab) = (b/c) × ((2c² - a² + c² - b²)/ (2b² - a² - b² +c²)) = (b/c) × ((c² - a² + b²)/ (b² - a² +c²)) = b/c ... (1) Now, use the sine rule, sina/a = sinb/b = sinc/c = i/k So, a = asina, b = ksinb and c = ksinc Substitute the value of b and c in equation (1). b/c = ksinb / ksinc = sinb / sinc Hence proved.

Answer»

Solution:

Given: ΔABC

Prove that sinb/sinc = c- acosb/b-acosc

Use cosine formulas:

cosb = (a²+c² - b²)/2ac and cosc = (a² - c² + b² )/ 2ab

RHS:

c- acosb/b-acosc = (c- a (a²+c² - b²)/2ac)/ (b - a (a² - c² + b² )/ 2ab)

= (b/c) × ((2c² - a² + c² - b²)/ (2b² - a² - b² +c²))

= (b/c) × ((c² - a² + b²)/ (b² - a² +c²))

= b/c ... (1)

Now, use the sine rule,

sina/a = sinb/b = sinc/c = i/k

So, a = asina, b = ksinb and c = ksinc

Substitute the value of b and c in equation (1).

b/c = ksinb / ksinc

= sinb / sinc

Hence proved.

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