Question
In $\triangle \text{ABC}, \ \angle \text{B} =60^\circ, \angle \text{C}=40^\circ,\text{AL}\bot \text{BC}$ and $AD$ bisects $\angle \text{A}$ such that $L$ and $D$ lie on side $BC$. Find $\angle \text{LAD}$
✓
Answer
We know that the sum of all angles of a triangle is $180^\circ $
Therefore, for $\triangle \text{ABC},$ we can say that:
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$
Or,
$\angle \text{A}+60^\circ+40^\circ=180^\circ$
$\Rightarrow\angle \text{A} =80^\circ$
$\angle \text{DAC}=\frac{\angle \text{A}}{2}$ $(\because $ AD bisects $\angle \text{A})$
$\Rightarrow \angle \text{DAC}=\frac{80^\circ}{2}=40^\circ$
If we use the aboce logic on $\triangle \text{ADC},$ we can say that:
$\angle \text{ADC}+\angle \text{DCA}+\angle \text{DAC}=180^\circ$ $($Sum of all angles of $\triangle \text{ADC})$
$\angle \text{ADC}+40^\circ+40^\circ=180^\circ$
$\angle \text{ADC}=180^\circ-80^\circ=100^\circ$
$\angle \text{ADC}=\angle \text{ALD}+\angle \text{LAD}$ (Exterior angle is equal to the sum of two interior opposite angles)
$100^\circ=90^\circ+\angle \text{LAD}$ (AL perpendicular to $BC)$
$\angle \text{LAD}=90^\circ$
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