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Question Number 61347 by Tawa1 last updated on 01/Jun/19

Commented by mr W last updated on 02/Jun/19

x=20°, see also Q43728.

$${x}=\mathrm{20}°,\:{see}\:{also}\:{Q}\mathrm{43728}. \\ $$

Answered by peter frank last updated on 01/Jun/19

x=40°

$${x}=\mathrm{40}° \\ $$

Answered by mr W last updated on 02/Jun/19

Commented by mr W last updated on 02/Jun/19

draw DF//AB  ΔDFG is equilateral, ⇒DF=GF  ΔAFC is isosceles, ⇒AF=CF  ΔAGC≡ΔCEA, ⇒AG=CE  AF−AG=CF−CE  ⇒GF=EF  ⇒ΔDFE is isosceles,  ⇒∠DEF=∠EDF=((180−∠DEF)/2)=((180−80)/2)=50°  ∠AEB=180−70−80=30°  x=∠DEF−∠AEB=50−30=20°

$${draw}\:{DF}//{AB} \\ $$$$\Delta{DFG}\:{is}\:{equilateral},\:\Rightarrow{DF}={GF} \\ $$$$\Delta{AFC}\:{is}\:{isosceles},\:\Rightarrow{AF}={CF} \\ $$$$\Delta{AGC}\equiv\Delta{CEA},\:\Rightarrow{AG}={CE} \\ $$$${AF}−{AG}={CF}−{CE} \\ $$$$\Rightarrow{GF}={EF} \\ $$$$\Rightarrow\Delta{DFE}\:{is}\:{isosceles}, \\ $$$$\Rightarrow\angle{DEF}=\angle{EDF}=\frac{\mathrm{180}−\angle{DEF}}{\mathrm{2}}=\frac{\mathrm{180}−\mathrm{80}}{\mathrm{2}}=\mathrm{50}° \\ $$$$\angle{AEB}=\mathrm{180}−\mathrm{70}−\mathrm{80}=\mathrm{30}° \\ $$$${x}=\angle{DEF}−\angle{AEB}=\mathrm{50}−\mathrm{30}=\mathrm{20}° \\ $$

Commented by otchereabdullai@gmail.com last updated on 11/Apr/20

wow

$$\mathrm{wow} \\ $$

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