MCQ
In a rhombus $\text{ABCD},$ if $\angle\text{ACB}=40^\circ,$ then $\angle\text{ADB}=$
- A$70^\circ$
- B$45^\circ$
- ✓$50^\circ$
- D$60^\circ$
✓
Answer
Correct option: C.
$50^\circ$
Consider $\triangle\text{AOD} \ \ \ \triangle\text{COB}$
$\angle\text{AOD}=\angle\text{COB}=90^\circ$
$AD = BC\ ($Sides of Rhombus$)$
$AO = CO\ ($Diagonals bisects each other$)$
So by $\text{RHS}$ property, $\triangle\text{AOD}\cong\triangle\text{COB}$
$\Rightarrow\angle\text{OAD}=\angle\text{OCB}=40^\circ$
$\angle\text{ADB}=\angle\text{ADO}$
$=180^\circ-90^\circ-40^\circ=50^\circ$
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