Question

Question 1
1 Point
Consider any triangle ABC with sides a=BC, b=CA & c=AB cosB/cosC =


option 1- (c-b.cosA)+(b-c.cosA)


option 2- (c-b.cosA)(b-c.cosA)


option 3- (c-b.cosA)/(b-c)


option 4- (c-b.cosA)/(b-c.cosA)

Answer 1

Answer:

Step-(b+c)cosA+(c+a)cosB+(a+b)cosC

⇒  bcosA+ccosA+ccosB+acosB+acosC+bcosC

⇒  (bcosC+ccosB)+(ccosA+acosC)+(acosB+bcosA)  ----( 1 )

Using projection formula,

a=(bcosC+ccosB)

b=(ccosA+acosC)

c=(acosB+bcosA)

Substituting above values in ( 1 ) we get,

⇒  a+b+c

∴   (b+c)cosA+(c+a)cosB+(a+b)cosC=a+b+c

by-step explanation: