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Question
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
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Solution
∠BAC = 180° - ∠BAD = 180° - 137° = 43°
∠ABC = 180° - ∠ABE = 180° - 106° = 74°
Thus, in ΔABC,
∠ACB = 180° - ∠BAC - ∠ABC
⇒ ∠ACB = 180° - 43° - 74° = 63°
Now, ∠ABC = ∠OBC + ∠ABO
⇒ ∠ABC = 2∠OBC ....( OB is biosector of ∠ABC )
⇒ 74° = 2∠OBC
⇒ ∠OBC = 37°
Similarly,
∠ACB = ∠OCB + ∠ACO
⇒ ∠ACB = 2∠OCB ...( OC is bisector of ACB )
⇒ 63° = 2∠OCB
⇒ ∠OCB = 31.5°
Now, in ΔBOC,
∠BOC = 180° - ∠OBC - ∠OCB
⇒ ∠BOC = 180° - 37° - 31.5°
⇒ ∠BOC = 111.5°
Since, ∠BOC> ∠OBC > ∠OCB, we have
BC > OC > OB
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