Question
In $\triangle \text{ABC},\angle \text{A}=100^\circ,$ AD bisects $\angle \text{A}$ and $\text{AD}\bot\text{BC}.$ Find $\angle \text{B}$
✓
Answer
Consider $\triangle \text{ABD}$$\angle \text{BAD}=\frac{100}{2}$ $($AD bisects $\angle \text{A})$
$\angle \text{BAD}=50^\circ$
$\angle \text{ADB}=90^\circ$ (AD perpendicular to BC)
We know that the sum of all three angles of a triangle is 180°
Thus,
$\angle \text{ABD}+\angle \text{BAD}+\angle \text{ADB}=180^\circ$ $($Sum of angles of $\triangle \text{ABD})$
Or,
$\angle \text{ABD}+50^\circ+90^\circ=180^\circ$
$\angle \text{ABD}=180^\circ-140^\circ$
$\angle \text{ABD}=40^\circ$
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