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Discussion Response Post to: sol of triangles

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Priyesh (1601)

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tanA+2tanB

=sinA/cosA+2sinB/cosB

sinAcosB+2sinBcosA/cosAcosB=0

=>so, we have to prove sinAcosB+2sinBcosA=0

now AD = (a/2)sinB (see fig.)

cosB = 2c/a

sinAcosB =asinB/b * 2c/a = 2 * (c/b) * sinB = 2 * sinC ------------------------1

now cosA = cos (90 + Q) = -sinQ = -(a/2sinC divided by a/2sinB ) (from fig.)

= -sinC/sinB

therefore 2sinBcosA = 2 sinB * -(sinC/sinB) = - 2* sinC -----------------2

from 1 & 2

sinAcosB+2sinBcosA=0

hence proved

Click on the figure to enlarge

"Imagination is more important than knowledge."

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