Question

In the figure given below, $\angle DAB={100}^{\circ }and\angle DBC={30}^{\circ }$. Find$\angle BDC$.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${100}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is D ${50}^{\circ }$

Consider $△ADBand△BDC,$
BD = BD..... (common side)
AD = DC..... (given)
BA = BC..... (given)
$\Rightarrow △ADB$ $\cong$ $△BDC$ (By SSS criteria)

Consider $△DBC,$
$\Rightarrow \angle DCB=\angle DAB={100}^{\circ }$....(congruent parts of congruent triangle)
$\angle DBC=30°$....... (given)

$\angle DCB+\angle CBD+\angle BDC=180°$ ....(Angle sum property of triangle)
$\Rightarrow 100°+30°+\angle BDC=180°$
$\Rightarrow 130°+\angle BDC=180°$
$\Rightarrow$$\angle BDC=180°-130°$
$\Rightarrow$$\angle BDC=50°$