sin2A+sin2B+sin2C =4sinAsinBsinC andcosA+cosB+cosC-1 =4sin.(A)/(2)sin.(B)/(2)sin.(C)/(2) Hence, (4sinAsinBsinC)/(4sin.(A)/(2)sin.(B)/(2)sin.(C)/(2)) =8cos.(A)/(2)cos.(B)/(2)cos.(C)/(2)
In any triangle ABC, prove the projection formulaa=bcosC+osB using vector method. View Solution In any triangle ABC , prove that following: bcosB+c cos C=acos(B−C) View Solution Knowledge Check In any triangle ABC, sinA−cosB=cosC, then angle B is Aπ2 Bπ3 Cπ4 Dπ6Submit...
cosθ=0সমীকরণের সাধারণ সমাধান হয়― Sine formula proof :Sine Rule The ratio of sine of angle of incidence to the sine of angle of refraction. f(x)=x3+ax2−bx+4অপেক্ষকের ক্ষেত্র...