Question : Minor arc $BC$ subtends $\angle BAC$ and $\angle BDC$ at points $A$ and $D$, respectively, on the circumference of the major sector of the circle with centre $O$. What is the value (in degrees) of $(\angle ABC+\angle ACB)$, if $\angle BDC=73^{\circ}?$
Option 1: $117^{\circ}$
Option 2: $107^{\circ}$
Option 3: $103^{\circ}$
Option 4: $113^{\circ}$
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Correct Answer: $107^{\circ}$
Solution :
Given: Minor arc $BC$ subtends $\angle BAC$ and $\angle BDC$ at points $A$ and $D$.
$\angle BDC=73^{\circ}$
So, $\angle BAC=73^{\circ}$ (common base)
We know that in $\triangle ABC$,
$\angle ABC+\angle BAC+\angle ACB=180^{\circ}$
Putting the value of $\angle BAC=73^{\circ}$, we have,
⇒ $\angle ABC+73^{\circ}+\angle ACB=180^{\circ}$
⇒ $\angle ABC+\angle ACB=180^{\circ}–73^{\circ}$
⇒ $\angle ABC+\angle ACB=107^{\circ}$
Hence, the correct answer is $107^{\circ}$.
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