Circles [NEW] — Maths STD 9 — Question

We have, $\angle\text{B}=70^\circ$

Since, ABCD is a cyclic quadrilateral

Then, $​​\angle\text{B}+\angle\text{D}=180^\circ$

$\Rightarrow70^\circ+\angle\text{D}=180^\circ$

$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$

Since, AB || DC

Then, $\angle\text{B}+\angle\text{C}=180^\circ$ [Co-interior angles]

$\Rightarrow70^\circ+\angle\text{C}=180^\circ$

$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$

Now, $\angle\text{A} +\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]

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

$\Rightarrow\angle\text{A}=180^\circ-110^\circ=70^\circ$