in triangle ABC show that abc/2∆ = r1 + r2 / 1+cosC
Answers
EXPLANATION.
In ΔABC.
Show that,
⇒ abc/2Δ = (r₁ + r₂)/(1 + cos c).
As we know that,
Formula of :
⇒ R = (abc)/4Δ.
⇒ r₁ = 4R sin(A/2) cos(B/2) cos(C/2).
⇒ r₂ = 4R sin(B/2) cos(A/2) cos(C/2).
Using this formula in the equation, we get.
From L.H.S.
⇒ abc/2Δ.
⇒ abc = 4RΔ.
⇒ (4RΔ)/(2Δ) = 2R.
From R.H.S.
⇒ (r₁ + r₂)/(1 + cos c).
⇒ [4R sin(A/2) cos(B/2) cos(C/2) + 4R sin(B/2) cos(A/2) cos(C/2)]/(1 + cos c).
⇒ 4Rcos(C/2)[sin(A/2) cos(B/2) + sin(B/2) cos(A/2)]/(1 + cos c).
⇒ 4Rcos(C/2)[sin(A + B)/2]/(1 + cos c).
⇒ 4Rcos(C/2)[cos(C/2)]/(1 + cos c).
⇒ [4Rcos²(C/2)]/(1 + cos c).
⇒ 2R[2cos²(C/2)]/(1 + cos c).
As we know that,
Formula of :
⇒ 1 + cos x = 2cos²(x/2).
Using this formula in the equation, we get.
⇒ 2R(1 + cos c)/(1 + cos c).
⇒ 2R.
Hence Proved.
Question
in triangle ABC show that abc/2∆ = r1 + r2 / 1+cosC
To Prove
abc/2∆ = r1 + r2 / 1+cosC
Answer
see in attachment
