From sine law, we have, a/sinA = b/sinb = c/sinC = k => a = ksinA, b = ksinB, c = ksinC Now, L.H.S. = acosA+bcosB+c cosC =ksinAcosA+ksinBcosB+ksinCcosC =k/2[2sinAcosA+2sinBcosB+2sinCcosC] =k/2[sin2A+sin2B+sin2C] Now, in any traingle ABC, sin2A+s
In any triangle A B C , prove that following: \ \ acosA+bcosB+c\ cos C=2bsinAsinC=2csinAsinB
a=2RsinA,b=2RsinB,c=2RsinC 4. Substituting Values in Area Formula: Now, substituting the values of a and b into the area formula: Δ=12(2RsinA)(2RsinB)sinC Simplifying this, we get: Δ=2R2sinAsinBsinC 5. Expressing Δ in Terms of R: Now, we can express Δ in terms of R...