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Unformatted text preview: triangle . Using the halftriangle, we find: h1 2 cos * 1 = That is θ = 2 cos1 (1h) The surface of the triangle is equal to twice the surface of the half right triangle. 2 h)(1 L triange half of Surface = For right triangle: L 2 + (1h) 2 = 1 2 that is 2 h)(11 L = So, finally A = S 1 = S triangle = 2 1h)(11 h)(1h)(1 cos 2 h)(1 L 22 = Realizing that cos1 (1h) = π/2 – sin1 (1h), we obtain: h)(1 sinh)(11 h)(12 A 12 = A θ L h...
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This note was uploaded on 08/05/2008 for the course CHE 100 taught by Professor Croiset during the Fall '07 term at Waterloo.
 Fall '07
 CROISET
 Chemical Engineering

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