Question

Let ABCD be a kite with AC and BD as diagonals and $\angle ABC$ = ${30}^{\circ }$. Then $\angle ACB$=.

• A. 70°
• B. 75°
• C. 80°
• D. 85°

Answer 1

The correct option is B 75°

In $\Delta ABD$ and $\Delta CBD$
$AD=CD$ and $AB=CB$ [Adjacent sides of a kite]
$BD=BD$ [common]
$\Delta ABD\cong \Delta CBD$ [SSS congruence condition]
Also, $\angle ADB$ = $\angle CDB$ [C.P.C.T]
and $\angle ABD$ = $\angle CBD$ [C.P.C.T]
$\angle ABC={30}^{\circ }$ [Given]
$⟹\angle ABD+\angle CBD={30}^{\circ }$
$⟹2\angle CBD=30$
$⟹\angle CBD=\frac{30}{2}$
$⟹\angle CBD={15}^{\circ }$

In $\Delta OBC$,
$\angle BOC$ = ${90}^{\circ }$ [diagonals intersect at ${90}^{\circ }$ ]
$\angle BOC+\angle OBC+\angle OCB={180}^{\circ }$ (Angle sum property)
${90}^{\circ }+{15}^{\circ }+x={180}^{\circ }$
$x={180}^{\circ }-{105}^{\circ }$
$x={75}^{\circ }$