Question
In a rhombus ABCD, if $\angle\text{ACB}=40^\circ,$ then $\angle\text{ADB}=$
- 70°
- 45°
- 50°
- 60°
✓
Answer
- 50°
Solution:
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 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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