Question

In the given figure, chord $BE=BD$, $\angle CBD={33}^o$ and $OB\perp AC$, then $x+y-z$ is equal to:

• A. ${230}^o$
• B. ${237}^o$
• C. ${337}^o$
• D. None of these

Answer 1

Answer: (D)

None of these

In $\Delta BED$,

$BE=BD$ (given)

$\angle BED=\angle BDE$ (opposite angles are equals).

Given, $x=\angle BDE$.

Also, $\angle OBC={90}^{\circ }$ (given)

$\angle OBD+\angle CBD={90}^{\circ }$

$⟹$ $\angle OBD+{33}^{\circ }={90}^{\circ }$ ....[Since $\angle CBD={33}^o$]

$⟹$$\angle OBD={90}^o-{33}^o={57}^{\circ }$.

Now, $x+z={180}^{\circ }$ (opposite angles of a cyclic quadrilateral)

$⟹$$z={180}^{\circ }-x$ ----- (i).

In $\Delta BED$,

$\angle BED+\angle BDE+\angle EBD={180}^{\circ }$ (By angle sum property of triangle)

$⟹$$x+y+{57}^{\circ }={180}^{\circ }$

$⟹$$x+y+{57}^{\circ }={180}^{\circ }-x$

$⟹$$x+y+{57}^{\circ }=z$ .....(from (i))

$⟹$$x+y-z=-{57}^{\circ }$.

Hence, option $D$ is correct.