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Question
In ∆DEF, ∠F = 48°, ∠E = 68° and bisector of ∠D meets FE at G. Find ∠FGD
Solution
Given ∠F = 48°
∠E = 68°
In ∆DEF,
∠D + ∠F + ∠E = 180° ...[By angle sum property]
∠D + 68° + 68° = 180°
∠D + 116° = 180°
∠D = 180° – 116°
= 64°
Since DG is the angular bisector of ∠D
∠FDG = ∠GDE
Also ∠FDG + ∠GDE = ∠D
2 ∠FDG = 64°
2 ∠FDG = 64°
∠FDG = `(64^circ)/2` = 32°
∠FDG = 32°
In ∆FDG,
∠FDG + ∠GFD = 180° ...[By angle sum property of triangles]
32° + ∠FDG + 48° = 180°
∠FDG + 80° = 180°
∠FDG = 180° – 80°
∠FDG = 100°
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