Question

In the figure, given below, AD = BC,

$\angle BAC={30}^{\circ }$ and

$\angle CBD={70}^{\circ }$. Find: $\angle BCA$

Answer 1

In the figure, ABCD is a cyclic quadrilateral AC and BD are its diagonals $\angle BAC={30}^{\circ }$ and $\angle CBD={70}^{\circ }$

$\angle CAD=\angle CBD={70}^{\circ }$

Similarly $\angle BAC=\angle BDC={30}^{\circ }$

$\therefore \angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$

$\because$ AD = BC (given)

$\therefore \angle ABCD$ is an isosceles trapezium

and AB || DC

$\angle BAC=\angle DCA$ (alternate angle)

$\Rightarrow DCA={30}^{\circ }$

$\therefore ABD=\angle DAC={30}^{\circ }$

(Angles in the same segment)

$\therefore \angle BCA=\angle BCD-\angle DAC={80}^{\circ }-{30}^{\circ }={50}^{\circ }$