A,b,c are sides of a triangle prove that 1/b+c-a + 1/c+a-b
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Sol: Triangle ABC is right angled at C.Let BC = a, CA = b, AB = c. (i) Area of ΔABC = 1/2 × Base × Height = 1/2 × BC × AC = 1/2 ab Area of ΔABC = 1/2 × Base × Height = 1/2 × AB × CD = 1/2 cp ⇒ 1/2 ab = 1/2 cp ⇒ ab = cp Hence proved. (ii) In right angled triangle ABC,AB2 = BC2 + AC2 c2 = a2 + b2 (ab/p)2 = a2+ b2 a2b2/p2 = a2 + b2 -------- From proof (1)1/p2 = (a2 + b2) / a2b2 1/p2 = (a2 / a2b2 + b2/ a2b2) 1/p2 = (1/b2 + 1/a2) 1/p2 = (1/a2+ 1/b2) Hence proved.
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It's easy you can do it by using Trigonometry..........
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