Question

In a trapezium $ABCD$, $AB||DC,AB=AD,\angle ADC={64}^{\circ }$ and $\angle BCD={54}^{\circ }$. Find $\angle DBC$.

• A. ${64}^{\circ }$
• B. ${72}^{\circ }$
• C. ${94}^{\circ }$
• D. ${116}^{\circ }$

Answer 1

The correct option is C ${94}^{\circ }$

In trapezium, $ABCD$, $AB\parallel CD$
$\angle DAB+\angle ADB=180$ (Angles in parallel lines)
$\angle DAB=180-64=116$
In $△ABD$
$AB=AD$ (Given)
$\angle ADB=\angle ABD$ (Isoscles triangle property)
Sum of angles of triangle ABD = 180
$\angle ADB+\angle ABD+\angle DAB=180$
$2\angle ADB=180-116$
$\angle ADB={32}^{\circ }$
Now, given, $\angle D=64$
$\angle ADB+\angle BDC=64$
$\angle BDC=64-32={32}^{\circ }$
In $△DBC$
Sum of angles $=180$
$\angle DBC+\angle BDC+\angle C=180$
$32+\angle BDC+54=180$
$\angle BDC=180-86$
$\angle BDC={94}^{\circ }$