Question

In Fig. 5, O is the centre of the circle. The angle by the arc BCD at the centre is 140°, BC is produced to P. Find ∠DCP.

Answer 1

We have,

$\angle \mathrm{BOD}=140°$

Using the property of circles, the angle subtended by a chord at the center is twice the angle subtended by it on the remaining part of the circle, we get

$$\begin{gathered} \angle \mathrm{BAD}=\frac{1}{2}\angle \mathrm{BOD} \\ =\frac{1}{2}\times 140° \\ =70° \end{gathered}$$

Also, in cyclic quadrilateral ABCD,

As, the opposite angles are supplementary.

So,

$$\begin{gathered} \angle \mathrm{DCB}=180°-\angle \mathrm{BAD} \\ =180°-70° \\ =110° \end{gathered}$$

Now,

$$\begin{gathered} \angle \mathrm{DCP}+\angle \mathrm{DCB}=180°\left(\mathrm{Linear}\mathrm{pair}\right) \\ \Rightarrow \angle \mathrm{DCP}=180°-\angle \mathrm{DCB} \\ \Rightarrow \angle \mathrm{DCP}=180°-110° \\ \therefore \angle \mathrm{DCP}=70° \end{gathered}$$