Math, asked by tsha1t5elnivisankish, 1 year ago


ABCDEF IS A REGULAR HEXAGON . FIND EACH ANGLE OF TRIANGLE ABE

Answers

Answered by Manjula29
4
AB = CB (sides of regular hexagon ABCDEF) AF = CD (sides of regular hexagon ABCDEF) ∠BAF = ∠BCD (angle of regular hexagon ABCDEF) ∴∆ABF ≅ ∆CBD (SAS congruency) ⇒ BF = BD (CPCT) ... (6) And ∠ABF = ∠CBD ⇒ ∠CBD = 30° ... (7) In ∆BEF and ∆BED EF = ED (sides of regular hexagon ABCDEF) BF = BD [from (6)] BE = BE (common) ∴∆BEF ≅ ∆BED (SSS congruency) By CPCT ∠EBF = ∠EBD ... (8) Now ∠ABC = 120° [using (1)] ⇒∠ABF + ∠EBF + ∠EBD + ∠CBD = 120° ⇒ 30° + ∠EBF + ∠EBF + 30° = 120° [using (3), (6), (7) and (8)] ⇒ 2∠EBF = 120° – 60° = 60° ⇒ ∠EBF = 30° Now using angle sum property for ∆EBF ∠EBF + ∠BFE + ∠BEF = 180° ⇒30° + 90° + ∠BEF = 180° ∠BEF = 180° – 120° = 60°
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