Question

in any ∆ABC, prove that. acosA + bcosB + ccosC = 8∆^2/abc​

Answer 1

Answer:

MARK AS BRAINLIEST ANSWER.

Step-by-step explanation:

→ By using sine rule, we can obtain values of a, b and c and then by substituting these values in L.H.S. we can prove this.

→ a/sinA = b/sinB = c/sinC = k(let)[by sine rule]

→ Then, a ksinA, b = ksinB and c = ksinC

→ Now, acosA + bcosB + ccosC

→ ksinAcosA + ksinBcosB + ksinCsinC

→ k/2[sin2AsinBsinC] = 2k[sinAsinBsinC

→ 2asinBsinC = 2a.(2∆/ac)(2∆/ab)

→ ∴ ∆ = 1/2absinC = 1/2acsinB ∴ sinB = 2∆/bc, sinC = 2∆/ab]

→ 8∆2/abc = R.H.S-