The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

Given that the measures of the exterior angles is 104° and 136°. 

ABC is a triangle and base BC is extended to both sides, 
By linear pair axiom 1 we get, 
$\angle \mathrm{ABD}+\angle \mathrm{ABC}=180°\dots \dots \left(1\right)$ 
Substituting the value of $\angle \text{ABD }=\text{ 1}0\text{4°}$ in equation 1,
$104°+\angle \mathrm{ABC}=180°$ 
Subtracting 104° on RHS, 
$\angle \mathrm{ABC}=180°-104°$
$\Rightarrow \mathbf{\angle }\mathbf{ABC}\mathbf{=}\mathbf{76}\mathbf{°}$ 
By linear pair axiom 1, 
$\angle \mathrm{ACE}+\angle \mathrm{ACB}=180°\dots \dots \left(2\right)$ 
Substituting the value of $\angle \text{ACE }=\text{ 136}°$ in equation 2,
$136°+\angle \mathrm{ACB}=180°$ 
Subtracting $\text{136}°$ from 180°, 
$\angle \mathrm{ACB}=180°-136°$
$\Rightarrow \mathbf{\angle }\mathbf{ACB}\mathbf{=}\mathbf{44}\mathbf{°}$ 
The sum of the three angles present in a triangle,
$\angle \mathrm{A}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180°\dots \dots \left(3\right)$ 
Substituting the value of $\angle \text{ABC} =\text{ 76}°$ and $\angle \text{ACB} =\text{ 44}°$ in equation 3, 
$\Rightarrow \angle \mathrm{A}+76°+44°=180°$
$\Rightarrow \angle \mathrm{A}+120°=180°$
$\Rightarrow \angle \mathrm{A}=180°-120°$
$\Rightarrow \mathbf{\angle }\mathbf{A}\mathbf{=}\mathbf{60}\mathbf{°}$
Thus, the three angles of the triangle are $\mathbf{\text{6}}\mathbf{0}\mathbf{°}\mathbf{,}\mathbf{\text{ 76}}\mathbf{°}$ and $\mathbf{\text{44}}\mathbf{°}$.