Question
In the given figure, $AD$ divides $\angle\text{BAC}$ in the ratio $1 : 3$ and $AB = DB$. Determine the value of $x$.
✓
Answer
The angle $\angle\text{BAC}$ is divided by $AD$ in the ratio $1 : 3$.
Let $\angle\text{BAD}$ and $\angle\text{DAC}$ be $y$ and $3y$, respectively.
As $BAE$ is a straight line, $\angle\text{BAC}=\angle\text{CAE}=180^\circ $ [linear pair]
$\Rightarrow\angle\text{BAD}+\angle\text{DAC}+\angle\text{CAE}=180^\circ$
$\Rightarrow\text{y}+3\text{y}+108^\circ=180^\circ$
$\Rightarrow4\text{y}=180^\circ-108^\circ=72^\circ$
$\Rightarrow\text{y}=\frac{72^\circ}{4}=18^\circ$ Now, in $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$
$\text{y}+\text{x}+4\text{y}=180^\circ$
$[$ Since, $\angle\text{ABC}=\angle\text{BAD}$ (given $AD = DB$) and $\angle\text{BAC}=\text{y}+3\text{y}=4\text{y}]$
$\Rightarrow5\text{y}+\text{x}=180$
$\Rightarrow5\times18+\text{x}=180$
$\Rightarrow90+\text{x}=180$
$\Rightarrow\text{x}=180-90=90$
Let $\angle\text{BAD}$ and $\angle\text{DAC}$ be $y$ and $3y$, respectively.
As $BAE$ is a straight line, $\angle\text{BAC}=\angle\text{CAE}=180^\circ $ [linear pair]
$\Rightarrow\angle\text{BAD}+\angle\text{DAC}+\angle\text{CAE}=180^\circ$
$\Rightarrow\text{y}+3\text{y}+108^\circ=180^\circ$
$\Rightarrow4\text{y}=180^\circ-108^\circ=72^\circ$
$\Rightarrow\text{y}=\frac{72^\circ}{4}=18^\circ$ Now, in $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$
$\text{y}+\text{x}+4\text{y}=180^\circ$
$[$ Since, $\angle\text{ABC}=\angle\text{BAD}$ (given $AD = DB$) and $\angle\text{BAC}=\text{y}+3\text{y}=4\text{y}]$
$\Rightarrow5\text{y}+\text{x}=180$
$\Rightarrow5\times18+\text{x}=180$
$\Rightarrow90+\text{x}=180$
$\Rightarrow\text{x}=180-90=90$
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