Question
From the following figure find ;$(i)\  x\ (ii) \ \angle ABC\ (iii)\  \angle \ ACD$

Answer

$(i)$ In Quadrilateral $ABCD$,

$x+4 x+3 x+4 x+48^{\circ}=360^{\circ}$
$ 12 x=360^{\circ}-48^{\circ}$
$ 12 x=312$
$ x=\frac{312}{12}=26^{\circ}$
$(ii) \angle \mathrm{ABC}=4 \mathrm{x}$
$4 \times 26=104^{\circ}$
$(iii) \angle \mathrm{ACD}=180^{\circ}-4 \mathrm{x}-48^{\circ}$
$=180^{\circ}-4 \times 26^{\circ}-48^{\circ}$
$=180^{\circ}-104^{\circ}-48^{\circ}$
$ =180^{\circ}-152^{\circ}=28^{\circ}$

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