Question

$AOD$ is a diameter of the circle with centre $O.$ Given that $\angle BDA={18}^{\circ }$ and $\angle BDC={38}^{\circ }.$ Find $\angle BCD$.

• A. ${100}^{\circ }$
• B. ${108}^{\circ }$
• C. ${126}^{\circ }$
• D. ${152}^{\circ }$

Answer 1

Answer: (B)

${108}^{\circ }$

Given, $AD$ is a diameter.

So, $\angle ABD={90}^{\circ }$ (Angle in a semi circle)

Now, opposite angles of a cyclic quadrilateral are supplementary.

So,

$\angle ABC+\angle ADC={180}^{\circ }$

$\Rightarrow (\angle ABD+\angle DBC)+(\angle ADB+\angle BDC)={180}^{\circ }$

$\Rightarrow ({90}^{\circ }+\angle DBC)+({18}^{\circ }+{38}^{\circ })={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle DBC+{56}^{\circ }={180}^{\circ }$

$\Rightarrow \angle DBC+{146}^{\circ }={180}^{\circ }$

$\Rightarrow \angle DBC={34}^{\circ }$

Now, applying angle sum property In $△DBC$

$\angle DBC+\angle BCD+\angle BDC={180}^{\circ }$

$\Rightarrow {34}^{\circ }+\angle BCD+{38}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BCD+{72}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BCD={108}^{\circ }$

Hence, the measure of $\angle BCD$ is ${108}^{\circ }$