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Question
In ΔDEF, ∠D = 60°, ∠E = 70° and the bisectors of ∠E and ∠F meet at O. Find (i) ∠F (i) ∠EOF.
Solution
(i) As we know,
∠D + ∠E + ∠F = 180° ......[Angle sum property of a triangle]
⇒ 60° + 70° + ∠F = 180° ......[∵ ∠D = 60° and ∠E = 70°]
⇒ ∠F = 180° – 130°
⇒ ∠F = 50°
(ii) Now, as FO is the bisector of ∠F
So, ∠EFO = `(∠F)/2 = 50^circ/2` = 25°
And ∠OEF = `(∠E)/2 = 70^circ/2` = 35° ......[∵ ∠D = 60° and ∠E = 70°]
In ΔEOF, ∠EOF + ∠OEF + ∠OFE = 180° ......[Angle sum property of a triangle]
⇒ ∠EOF + 35° + 25° = 180°
⇒ ∠EOF = 180° – 60°
⇒ ∠EOF = 120°
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