hi sulekha,

due to shortage of time i would liketo give poper hints so that u can solve the problem urself .

wrtie cos A =(b²+ c² - a² )/2bc and write cos B ,cosC in hthe same manner and substitue in the left side equation and sinC/sinB=c/b and when take the given experssion adn solve , finally u'll get

 

(b-c)(a²-b²-c²)=0

 

from here either

b=c or                     a² = b² + c²    

isocless triangle or rightangled triangle

this can be written in the form

c cosA+2c cosC=b cos A+2b cos B

(b-c)(b^2+c^2-a^2)/2bc=c(a^2+b^2-c^2)/ab-b(a^2+c^2-b^2)/ac=

(c^2*a^2-c^4-b^2*a^2+b^4)/ab

(b^2+c^2-a^2)(b-c)/2=(c^2-b^2)(a^2-c^2-b^2)/a

and then u can equate b^2+c^2-a^2=0 and b-c=0

and in the cases u get one of them gives rt angled triangle and one of them gives isosceles triangle

please rate me