Question

In the figure given below, $\angle DAB={100}^{\circ }and\angle DBC={30}^{\circ }$. Find$\angle BDC$. [3 Marks]

Answer 1

Consider $△ADBand△BDC,$
BD = BD..... (common side)
AD = DC..... (given)
BA = BC..... (given)
$\Rightarrow △ADB$ $\cong$ $△BDC$ (By SSS criteria) [ 1 Mark]

Consider $△DBC,$
$\Rightarrow \angle DCB=\angle DAB={100}^{\circ }$....(congruent parts of congruent triangle)
$\angle DBC=30°$....... (given) [ 1 Mark]


$\angle DCB+\angle CBD+\angle BDC=180°$ ....(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°$ [1 mark]