MCQ
In $\triangle\text{ABC},$ $\angle\text{A}=100^{\circ},$ $AD$ bisects$\angle\text{A}$ and $AD$ $\bot$ $BC$. then, $\angle\text{B}$ is equal to:
- A$80^\circ $
- B$20^\circ $
- ✓$40^\circ $
- D$30^\circ $
✓
Answer
Correct option: C.
$40^\circ $
Given, $\angle\text{BAD}=\angle\text{DAC}=50^{\circ}$ $[\because\text{AD} \ \text{bisect}\angle\text{A} \ \text{and} \ \angle\text{A}=100^{\circ}]$
and $\angle\text{BDA}=\angle\text{ADC}=90^{\circ}$ $[\because\text{AD}\bot\text{BC}]$
Now, in $\triangle\text{ABD},$
$\angle\text{ABD}+\angle\text{BAD}+\angle\text{BDA}=180^{\circ}$ [angle sum property ofb triangle]
$\Rightarrow \ \angle\text{ABD}+50^{\circ}+90^{0}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}+140^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}=180^{\circ}-140^{\circ}$
$\Rightarrow \ \angle\text{ABD}=140^{\circ}$
and $\angle\text{BDA}=\angle\text{ADC}=90^{\circ}$ $[\because\text{AD}\bot\text{BC}]$
Now, in $\triangle\text{ABD},$
$\angle\text{ABD}+\angle\text{BAD}+\angle\text{BDA}=180^{\circ}$ [angle sum property ofb triangle]
$\Rightarrow \ \angle\text{ABD}+50^{\circ}+90^{0}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}+140^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}=180^{\circ}-140^{\circ}$
$\Rightarrow \ \angle\text{ABD}=140^{\circ}$
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