In ∆ABC it is given that ∠A=70°,∠B=52° ,BO and CO are are the

1.	In ∆ABC it is given that ∠A=70°,∠B=52° ,BO and CO are are the bisector of ∠B and ∠C respectively ,find ∠OCD and ∠BOC.
Answer» ━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$ In ∆ABC it is given that ∠A=70°,∠B=52° ,BO and CO are are the BISECTOR of ∠B and ∠C respectively ,find ∠OCD and ∠BOC. ━━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$ We have that the SUM of the ANGLES of a triangle is 180°. $\therefore \: \angle \: A + \angle \: B + \angle \: C = 180\degree$ $⇒ 70\degree + 52\degree + \angle \: C = 180\degree \\$ $⇒ \angle \: C = (180\degree - 122\degree) = 58\degree \\$ $\therefore \: \angle \: OCB = \frac{1}{2} \angle \: C = ( \frac{1}{2} \times 58\degree) = 29\degree \\$ $and \: \angle \: OBC = \frac{1}{2} \angle \: B = ( \frac{1}{2} \times 52\degree) = 26\degree \\$ In ∆BOC,we have $\angle \: OBC + \angle \: OCB + \angle \: BOC = 180\degree \\$ $⇒ 26\degree + 29\degree + \angle \: BOC = 180\degree \\$ $⇒ \angle \: BOC = (180\degree - 55\degree) = 125\degree \\$ $hence, \: \angle \: OCB = 29\degree \: and \: \angle \: BOC = 125\degree \\$ ━━━━━━━━━━━━━━━━━━━━━━━━━

In ∆ABC it is given that ∠A=70°,∠B=52° ,BO and CO are are the bisector of ∠B and ∠C respectively ,find ∠OCD and ∠BOC.

Answer»

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$\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$

In ∆ABC it is given that ∠A=70°,∠B=52° ,BO and CO are are the BISECTOR of ∠B and ∠C respectively ,find ∠OCD and ∠BOC.

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$\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$

We have that the SUM of the ANGLES of a triangle is 180°.

$\therefore \: \angle \: A + \angle \: B + \angle \: C = 180\degree$

$⇒ 70\degree + 52\degree + \angle \: C = 180\degree \\$

$⇒ \angle \: C = (180\degree - 122\degree) = 58\degree \\$

$\therefore \: \angle \: OCB = \frac{1}{2} \angle \: C = ( \frac{1}{2} \times 58\degree) = 29\degree \\$

$and \: \angle \: OBC = \frac{1}{2} \angle \: B = ( \frac{1}{2} \times 52\degree) = 26\degree \\$

In ∆BOC,we have

$\angle \: OBC + \angle \: OCB + \angle \: BOC = 180\degree \\$

$⇒ 26\degree + 29\degree + \angle \: BOC = 180\degree \\$

$⇒ \angle \: BOC = (180\degree - 55\degree) = 125\degree \\$

$hence, \: \angle \: OCB = 29\degree \: and \: \angle \: BOC = 125\degree \\$

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