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The exterior angles obtained on producing the base of a triangle both ways are $104^o$ and $136^o$. Find all the angles of the triangle.
Given:
The exterior angles obtained on producing the base of a triangle both ways are $104^o$ and $136^o$.
To do:
We have to find all the angles of the triangle.
Solution:
Let in a triangle $ABC$, base $BC$ is produced both ways to $D$ and $E$ respectively forming $\angle ABE = 104^o$ and $\angle ACD = 136^o$.
From the figure,
$\angle \mathrm{ABC}+\angle \mathrm{ABE}=180^{\circ}$
$\angle \mathrm{ABC}+104^{\circ}=180^{\circ}$
$\angle \mathrm{ABC}=180^{\circ}-104^{\circ}=76^{\circ}$
Similarly,
$\angle \mathrm{ACB}+\angle \mathrm{ACD}=180^{\circ}$
$\angle \mathrm{ACB}+136^{\circ}=180^{\circ}$
$\angle \mathrm{ACB}=180^{\circ}-136^{\circ}=44^{\circ}$
$\angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180^{\circ}$ (Angles sum property)
$76^{\circ}+44^{\circ}+\angle B A C=180^{\circ}$
$120^{\circ}+\angle B A C=180^{\circ}$
$\angle B A C=180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the angles of the triangle are $44^{\circ}, 60^{\circ}$ and $76^{\circ}$.
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