Question

if S is the circumcenter of ∆abc and R1,R2,R3 are the circum radii of ∆SBC,∆SCA,∆SAB then
a/R1+b/R2+C/R3 = ?​

Answer 1

Answer:

a/R1+b/R2+C/R3  = 4 tanA tanB tanC [∵A+B+C=$180^{0}$

Step-by-step explanation:

From the above question, S is the circumcenter of ∆abc and R1,R2,R3 are the circum radii of ∆SBC,∆SCA,∆SAB.

We know that Δ=abc/4R

Δ1 = $\frac{OB * OC * BC}{4R_{1} }$

BOD = $\frac{1}{2}$BOC = A

   In an acute angled triangle ABC,r+r1=r2+r3and∠B>π3, then b+2c<2a<2b+2c b+4c<4a<2b+4c b+4c<4a<4b+4c b+3c<3a<3b+3c

      r + r1 = r2 + r3

      r - r2 = r3 - r  

∆/s - ∆/s-b = ∆/s-c - ∆/s-a

     - b/s(s-b) = c-a/(s-a)(s-c)

      (s-a)(s-c)/ s(s-b) = a-c/b

tan B/2 = √(s-a)(s-c) / s(s-b)

   =  tan² B/2 = a-c/b*

      π/2 > B > π/3

      π/4 > B/2 > π/6

      1/3 < tan²(B/2) < 1

      1/3 < a-c/b < 1

      b/3 < a-c < b

      b < 3a-3c

      a-c < b

  = b + 3c < 3a

Here, $\frac{a}{r_{1} } + \frac{b}{r_{2} } + \frac{b}{r_{3} }$ = 4 (tan A + tan B + tan C)

a/R1+b/R2+C/R3  = 4 tanA tanB tanC [∵A+B+C=$180^{0}$]

For more related question : https://brainly.in/question/13918365

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