Question

In the given figure AB = BC, $\angle ABC={68}^o$ DA and DB are the tangents to the circle with centre O.
Calculate the measure of $\angle ADB$

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,

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

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

Now, $\angle AOB=2\angle ACB$ (Angle at centre is twice the angle at the circumference)

Therefore,

$\angle AOB=2\times {56}^0={112}^0$

Also, we know that $\angle AOB+\angle ADB={180}^0$, thus

$\angle AOB+\angle ADB={180}^0\Rightarrow {112}^0+\angle ADB={180}^0\Rightarrow \angle ADB={180}^0-{112}^0={68}^0$

Hence, $\angle ADB={68}^0$.