Question

In triangle abc prove that c-b cosa/b-c cosa=cosb/cosc

Answer 1

To prove --->

c - b CosA / b - c CosA = CosB / CosC

Proof --->

We know that from projection rule

c = b CosA + a CosB

b = a CosC + c CosA

Now

LHS= c - bCosA / b - c CosA

Putting value of c and b

b CosA + a CosB- b CosA

= ---------------------------------------

a CosC + c CosA - c CosA

+ bCosA and - bCosA cancel out from numerator and cCosA and - cCosA cancel out from denominator so we get

= a CosB / a CosC

a cancel out from numerator and denominator

= CosB / CosC = RHS

Additional information-->

(1) Sin rule

a /SinA = b / SinB = c / SinC

(2) Cos rule

CosA =( b² + c² - a²) / 2 bc

CosB = (a² + c² - b²) / 2 ac

CosC = (a² + b² - c²) / 2 ab