Question

In the given circle, with AB as diameter, find the value of 'x'. [2 Marks]

Answer 1

$\angle ACB={90}^{\circ }$ [Angle inscribed in a semicircle is a right angle] [1 Mark]

$\angle ACB=\angle BCD+\angle DCA$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow$$\angle BCD=\angle ACB-\angle DCA$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow$$\angle BCD={90}^{\circ }-{30}^{\circ }={60}^{\circ }$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow$$\angle DAB=\angle DCB={60}^{\circ }$ [Angles inscribed in the same arc are congruent]

$\therefore x=\angle DAB={60}^{\circ }$ [1 Mark]