Question

Suppose $AB=BC$, $\angle ABC={68}^o$ , $DA$ and $DB$ are the tangents to the circle with centre O, for the trianlge $ABC$
Calculate the measure of $\angle ACB$

Answer 1

It is given that $\angle ABC={68}^0$ and $AB=BC$.

In $△ABC$, the sum of three angles is ${180}^0$ that is:

$\angle CAB+\angle ABC+\angle ACB={180}^0$

But $\angle CAB=\angle ACB$ (Angle opposite to equal sides are equal)

Therefore,

$\angle ACB+\angle ACB+{68}^0={180}^0\Rightarrow 2\angle ACB+{68}^0={180}^0\Rightarrow 2\angle ACB={180}^0-{68}^0\Rightarrow 2\angle ACB={112}^0\Rightarrow \angle ACB=\frac{{112}^0}{2}={56}^0$

Hence, $\angle ACB={56}^0$.