Question

If $\angle BOC=80^{\circ}$ and $OA$ bisects $\angle BAC$, then find the value of $\angle ABO$ in degrees.

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Answer

Given: $\angle BOC = 80^\circ$ and $OA$ is a line that bisects $\angle BAC$.

We know that, an angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, we can write:

$$ \angle BOC = 2 \times \angle BAC $$

Substituting the given value:

$$ 80^\circ = 2 \times \angle BAC \ \therefore \angle BAC = 40^\circ $$

Since $OA$ bisects $\angle BAC$, we can conclude:

$$ \angle BAO = \frac{\angle BAC}{2} = \frac{40^\circ}{2} = 20^\circ $$

Now, consider the triangle $\triangle AOB$:

Since $OA = OB = r$ (radii of the circle), $\triangle AOB$ is isosceles, making the angles opposite the equal sides equal. Therefore:

$$ \angle OAB = \angle OBA $$

Given that $\angle BAO = 20^\circ$, we find:

$$ \angle ABO = 20^\circ $$

So, the value of $\angle ABO$ is 20 degrees.


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