Question

The given figure shows a circle through the points A, B, C, and D. If $\angle BAC={67}^{\circ }$, find $\angle DBC+\angle DCB$

• A. ${128}^{\circ }$
• B. ${134}^{\circ }$
• C. ${113}^{\circ }$
• D. ${67}^{\circ }$

Answer 1

The correct option is C

${113}^{\circ }$

Given, $\angle BAC={67}^{\circ }$
$\angle BDC=\angle BAC={67}^{\circ }$ [Angle in the same segment are equal]

In $\Delta DBC$

$\angle BDC+\angle DBC+\angle DCB={180}^{\circ }$ [Sum of angles of a triangle is ${180}^{\circ }$]

$\Rightarrow {67}^{\circ }+\angle DBC+\angle DCB={180}^{\circ }$

$\Rightarrow \angle DBC+\angle DCB={180}^{\circ }-{67}^{\circ }={113}^{\circ }$