Question Number 197717 by a.lgnaoui last updated on 27/Sep/23
$$\measuredangle\boldsymbol{\mathrm{A}}=\mathrm{62}\:\:\:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{43} \\ $$$$\boldsymbol{\mathrm{Determiner}}:\:\:\mathrm{a}=\measuredangle\boldsymbol{\mathrm{B}}\:\:\:\:\mathrm{c}=\measuredangle\boldsymbol{\mathrm{D}}\:\:\:\:\:\mathrm{b}=\measuredangle\boldsymbol{\mathrm{F}} \\ $$
Commented by a.lgnaoui last updated on 27/Sep/23
$$\measuredangle{EDF}=\mathrm{62}\:? \\ $$
Commented by a.lgnaoui last updated on 27/Sep/23
Commented by som(math1967) last updated on 27/Sep/23
$$\measuredangle{AEC}=\mathrm{180}−\left(\mathrm{62}+\mathrm{43}\right)=\mathrm{75} \\ $$$$\therefore\angle{ABD}=\mathrm{180}−\mathrm{75}=\mathrm{105} \\ $$$$\angle{EDF}=\mathrm{62} \\ $$$$\angle{EFD}=\mathrm{180}−\left(\mathrm{62}+\mathrm{105}\right)=\mathrm{13} \\ $$
Commented by som(math1967) last updated on 27/Sep/23
$${yes},\:\angle{BDE}+\angle{BAE}=\mathrm{180} \\ $$$$\angle{BDE}+\angle{EDF}=\mathrm{180} \\ $$$$\therefore\angle{BDE}+\angle{BAE}=\angle{BDE}+\angle{EDF} \\ $$$$\therefore\angle{BAE}=\angle{EDF}=\mathrm{62} \\ $$
Commented by a.lgnaoui last updated on 27/Sep/23
$$\mathrm{thanks} \\ $$