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The Bisects of Exterior Angle at B and C of δAbc Meet at O. If ∠A = X°, Then ∠Boc = - Mathematics

MCQ

The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =

Options

  • \[90^\circ + \frac{x^\circ }{2}\]
  • \[90^\circ - \frac{x^\circ }{2}\]

  • \[180^\circ + \frac{x^\circ }{2}\]

  • \[180^\circ - \frac{x^\circ }{2}\]

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Solution

In the given figure, bisects of exterior angles ∠Band ∠C meet at O and  ∠A = x°

We need to find  ext. ∠BOC

Now, according to the theorem, “if the sides AB and AC of a ΔABC are produced to P and Qrespectively and the bisectors of  ∠PBC and ∠QCB intersect at O, therefore, we get,

`∠BOC = 90^\circ - 1/2 ∠A`

Hence, in ΔABC

`∠BOC = 90^\circ - 1/2 ∠A`

`∠BOC = 90^\circ - 1/2 x`

Thus,

`∠BOC = 90^\circ - x/2`

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 11 Triangle and its Angles
Q 26 | Page 29
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