Question

in the given figure ab∥fg, abd and edf are straight lines and cbed is a rhombus the measure of ∠dfg-∠bde+∠cdf is equal to 1)40 2)60 3)70 4)20

Answer 1

Given : ab∥fg, abd and edf are straight lines and cbed is a rhombus

To Find : measure of ∠dfg-∠bde+∠cdf

Solution:

∠DFG = 115°

∠ABE = 105°

∠ABE + ∠DBE = 180°   Linear Pair

=>105° +  ∠DBE =   180°

=> ∠DBE =   75°

=> ∠BDE = ∠DBE  as  BE = DE sides of rhombus

=> ∠BDE  =   75°

∠BDC = ∠DBE = 75°   ( alternate angle as opposite sides of rhombus are parallel)

∠ADF = ∠DFG = 115°  as AB ||  FG

=> ∠ADF =  115°  

∠ADF = ∠DBE  + ∠CDF

=> 115°   = 75°  + ∠CDF

=> ∠CDF = 40°

∠DFG-∠BDE+∠CDF  = 115° - 75°  + 40°  = 80°

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