Question

ABCD is a rhombus. If $\angle BAC={38}^{\circ }$, find $\angle ADC$

• A. ${84}^o$
• B. ${104}^o$
• C. ${36}^o$
• D. ${68}^o$

Answer 1

Answer: (B)

Given is a rhombus ABCD & $\angle BAC={38}^o$.

We know that all sides of a rhombus are equal.

So AB=BC=CD=DA

In isosceles $△$ ABC,

$\angle BAC=\angle BCA={38}^o$[ Base angles are equal in isosceles triangle ]

Again in isosceles triangle ABC,

$\angle BAC+\angle BCA+\angle ABC={180}^o$

$\Rightarrow \angle ABC={180}^o-{38}^o-{38}^o$

$\Rightarrow \angle ABC={180}^o-{76}^o$

$\Rightarrow \angle ABC={104}^o$

We know that Opposite angles of a rhombus have equal measure.

So $\angle ABC=\angle ADC$

Therefore $\angle ADC={104}^o$