Find #m/_x+m/_y+m/_z# in the following figure?

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2 Answers
Jan 9, 2018

#m/_x+m/_y+m/_z=190^@#

Explanation:

As #m/_EBC=70^@#, #m/_ABE=180^@-70^@=110^@#

But #m/_ABD=150^@#

Hence #m/_x=150^@-110^@=40^@#

As #AE#||#BD#, #m/_x=m/_y# - (alternate angles)

Hence #m/_y=40^@#

Further as #BE#||#CD#, #m/_C=m/_ABE=110^@#

Hence #m/_z=110^@#

and #m/_x+m/_y+m/_z=40^@+40^@+110^@=190^@#

Jan 9, 2018

#color(blue)(X = 40^0, Y = 40^0, Z = 110^0#

#color(blue)(X + Y + Z = 40 + 40 + 110 = 190^0)#

Explanation:

#/_(ABE + /_(EBD) + /_(DBC) = 180^0# Eqn (1)

/_(ABE) = 150 - X#

#/_(DBC) = 70 - X#

Substituting values of #/_(ABE), /_(DBC)# in Eqn (1),

#150 - X + X + 70 - X = 180#

#color(red)X = 220 - 180 = color (red)40^0#

#/_(AEB) (Y), /_(EBD) (X)# are alternate angles.

Therefore #X = color(red)(Y = 40^0)#

#/_(EBC) + Z = 180# as the angles are supplementary.

Given #/_(EBC) = 70^0#

Therefore #color(red)Z = 180 - /_(EBC) = 180 - 70 = color (red)110^0#

#color(blue)(X + Y + Z = 40 + 40 + 110 = 190^0)#