# If angle BOC = 80 degrees and OA bisects angle BAC, then find the value of angle ABO in degrees.

## 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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