MCQ
Two chords $AB$ and $CD$ intersect at right angles. If, $\angle\text{BAC}=40^\circ,$ then $\angle\text{ABD}$ is equal to: 
- ✓$50^\circ$
- B$55^\circ$
- C$45^\circ$
- D$60^\circ$
✓
Answer
Correct option: A.
$50^\circ$
Let $AB$ and $CD$ intersect at $O.$ Then In the triangle $ACO,$
$\angle\text{A}+\angle\text{C}+\angle\text{D}=180^\circ$
$40^\circ+\angle\text{C}+90^\circ=180^\circ$
$130^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=50^\circ$
As per the theorem, Angles in the same segment of a circle are equal.
Hence, $\angle\text{ABD}=\angle\text{C}=50^\circ$
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