History of Hindu Mathematics: A Source Book v2.c17.s7.82
History of Hindu Mathematics: A Source Book v2.c17.s7.82
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v2.c17.s7.82
From (2) $x = (100 I)/r₁t₁, y = (100 I)/r₂t₂, z = (100 I)/r₃t₃.$ Substituting in (1), we get $100 I ( 1/r₁t₁ + 1/r₂t₂ + 1/r₃t₃ ) = c.$ Therefore $I = c/(100 ( 1/r₁t₁ + 1/r₂t₂ + 1/r₃t₃ )).$