Question

From the following figure find,
$\angle ACD$

Answer 1

In a quadrilateral $ABCD$

$\begin{array}{c}\Rightarrow \angle A+\angle B+\angle C+\angle D={360}^{\circ }\\ \Rightarrow 4x+x+{48}^{\circ }+3x+4x={360}^{\circ }\\ \Rightarrow 12x^{\circ }={360}^{\circ }-{48}^{\circ }\\ \Rightarrow 12x^{\circ }={312}^{\circ }\\ x^{\circ }={26}^{\circ }\\ In\Delta ACD,\angle D+\angle DAC+\angle ACD={180}^{\circ }\\ 4\times {26}^{\circ }+{48}^{\circ }+\angle ACD={180}^{\circ },\\ \angle ACD+{152}^{\circ }={180}^{\circ }\\ \angle ACD={28}^{\circ }\end{array}$