In , if =100^ ,AD bisects and AD . Then, = - Vidyadip

MCQ

In $\triangle\text{ABC},$ if $\angle\text{A}=100^\circ,\text{AD}$ bisects $\angle\text{A}$ and $\text{AD}\perp\text{BC}.$ Then, $\angle\text{B}=$

A
50º
B
90º
C
40º
D
100º

✓

Answer

40º
Solution:

$\text{AD}\perp\text{BC}$ and AD bisects $\angle\text{A}.$
$\Rightarrow\angle\text{BAD}=\angle\text{CAD}=50^\circ$
In Right $\triangle\text{ADB}$
$\angle\text{BAD}=50^\circ,\angle\text{ADB}=90^\circ$
Also sum of all interior angles = 180º
$\Rightarrow\angle\text{BAD}+\angle\text{ADB}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-50^\circ-90^\circ$
$\Rightarrow\angle\text{B}=40^\circ$

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