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Question

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A $AB = AC$

B $AB < AC$

C $AB > AC$

D $AB < BC$

Answer
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We have been provided in $\Delta ABC$, $\angle C = {110^ \circ} $. So, we need to show which of the above statement is true for that as we know the angle sum property of triangle which is that the sum of all the angles of a triangle is ${180^ \circ} $.

$\angle A + \angle B + \angle C = {180^ \circ} $

We have been given that $\angle C = {110^ \circ} $

Now we will be keeping the value of C in the above equation, from which we would get

$\angle A + \angle B + {110^ \circ} = {180^ \circ} $

$\angle A + \angle B = {70^ \circ} $

As we know the angle measuring less than ${90^ \circ} $ is called an acute angle. Since, here we have the sum of both the angles $\angle A$ and $\angle B$ is ${70^ \circ }$, which is less than ${90^ \circ }$, we can surely say that both angles $\angle A$and $\angle B$ must be acute angles.

Now we will be using the property of the triangle that is the side opposite to the greater angle is longer.

As $\angle C$ is the greatest so the side opposite to $\angle C$ is AB

So, we can say that the side $AB > AC$.