Question

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

विकल्प

• \[90^\circ + \frac{x^\circ }{2}\]

• \[90^\circ - \frac{x^\circ }{2}\]

• \[180^\circ + \frac{x^\circ }{2}\]

• \[180^\circ - \frac{x^\circ }{2}\]

Answer 1

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`