Question

In isosceles trapezoid, $\overline{AB}$ is parallel to $\overline{DC}$, $\angle BDC$ measures ${25}^{\circ }$ and $\angle BCA$ measures ${35}^{\circ }$ as shown in figure above. Find the measure of $\angle DBC$.

• A. ${85}^{\circ }$
• B. ${95}^{\circ }$
• C. ${105}^{\circ }$
• D. ${115}^{\circ }$
• E. ${125}^{\circ }$

Answer 1

The correct option is B ${95}^{\circ }$

Trapezoid is isosceles, so

$\angle ACD=\angle BDC={25}^{\circ }$

$\angle BCD=\angle ACD+\angle BCA$

$\angle BCD={25}^{\circ }+{35}^{\circ }={60}^{\circ }$

Sum of angle of $△BCD$ is ${180}^{\circ }$

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

$\Rightarrow \angle DBC={180}^{\circ }-{25}^{\circ }-{60}^{\circ }$

$\Rightarrow \angle DBC={95}^{\circ }$