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प्रश्न
The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =
पर्याय
- \[90^\circ + \frac{x^\circ }{2}\]
\[90^\circ - \frac{x^\circ }{2}\]
\[180^\circ + \frac{x^\circ }{2}\]
\[180^\circ - \frac{x^\circ }{2}\]
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उत्तर
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`
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