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in Co-ordinate geometry by (63.5k points)

If a, b, c are the sides of triangle ABC satisfying log(1 + c/a) + loga - logb - log2.

Also, a(1 - x2) + 2bx + c(1 + x2) = 0 has two equal roots. Find the value of sinA + sinB + sinC.

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Best answer

((a + c)/a)(a/b) = 2

a + c = 2b ... (i)

⇒ a(1 – x2) + 2bx + c(1 + x2) = 0

x2(c – a) + 2bx + c + a = 0

for equal roots b2 – c2 + a2 = 0

a2 + b2 = c2 ... (ii)

Hence it is right angle at ∠C

Putting value in equation (2) from equation (1)

⇒ a2 + b2 = (2b – a)2 ⇒ a2 + b2 = 4b2 + a2 – 4ab

⇒ 3b2 = 4ab ⇒ b = 4/3a; c = 5/4b ⇒ c = 5/3a

Hence a = 3k, b = 4k, c = 5k

sinA + sinB + sinC = 3/5 + 4/5 + 1 = 12/5

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