Question #02b52

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Answer 1 · 2017-03-17T04:12:55

see explanation.

Case 1.
enter image source here

#sinangleCAD=SinA=(CD)/(AC)#
#=> CD=AC*sinA=b*sinA# --- (1)

#sinangleCBD=SinB=(CD)/(BC)#
#=> CD=BC*SinB=a*sinB# ---(2)

(1) = (2), #=> b*SinA=a*SinB#

#=> SinA/a=sinB/b#

Case 2.
enter image source here

#sinangleCAD=SinA=(CD)/(AC)#
#=> CD=AC*sinA=b*sinA# --- (1)

#angleCBA=angleB#
#angleCBD=180-angleB#
#sinangleCBD=Sin(180-B)=sinB=(CD)/(BC)#,
#=> CD=BC*SinB=a*sinB# ------(2)

(1) = (2), #=> b*SinA=a*SinB#

#=> SinA/a=sinB/b#