MCQ
In $\triangle\text{ABC, }\text{BD}\perp\text{AC, }\angle\text{CAE} = 30^\circ$ and $\angle\text{CBD}=40^\circ.$ Then $\angle\text{AEB}=?$
- A$70^\circ$
- B$50^\circ$
- C$60^\circ$
- ✓$80^\circ$
✓
Answer
Correct option: D.
$80^\circ$
In $BDC$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\text{BD}\perp\text{AC}$
$\angle\text{BCD}=90^\circ,\angle\text{DBC}=40^\circ$
$90^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\angle\text{BCD}+130^\circ=180^\circ$
$\angle\text{BCD}=180^\circ-130^\circ$
$\angle\text{BCD}=50^\circ$
$\angle\text{AEB}=\angle\text{CAE}+\angle\text{C}$ (exterior angle)
$\angle\text{CAE}=30^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{AEB}=30^\circ+50^\circ$
$\angle\text{AEB}=80^\circ.$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\text{BD}\perp\text{AC}$
$\angle\text{BCD}=90^\circ,\angle\text{DBC}=40^\circ$
$90^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\angle\text{BCD}+130^\circ=180^\circ$
$\angle\text{BCD}=180^\circ-130^\circ$
$\angle\text{BCD}=50^\circ$
$\angle\text{AEB}=\angle\text{CAE}+\angle\text{C}$ (exterior angle)
$\angle\text{CAE}=30^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{AEB}=30^\circ+50^\circ$
$\angle\text{AEB}=80^\circ.$
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