Question 302580
<pre><font size = 4 color = "indigo"><b>
To avoid too many fractions, let's choose the congruent sides to have
measure 4 each and the base half that or 2, then we'll be able to 
take half the base without having a fraction for the measure of a side: 


{{{drawing(240,400,-.5,2.5,-.5,4.5,triangle(0,0,2,0,4*cos(1.318116072),4*sin(1.318116072)), locate(1,0,2), locate(.3,2,4), locate(1.6,2,4)

  )}}}

Draw a median from the vertex angle which is also the perpendicular
bisector as well as the bisector of the vertex angle.  That divides
the triangle into two congruent right triangles, each with a base
of measure 1:

{{{drawing(240,400,-.5,2.5,-.5,4.5,triangle(0,0,2,0,4*cos(1.318116072),4*sin(1.318116072)), locate(.5,0,1), locate(3.5,0,1), locate(.3,2,4), locate(1.6,2,4),
green(line(1,0,4*cos(1.318116072),4*sin(1.318116072))),
locate(.3,.3,theta),
red(arc(0,0,.6,-.6,0,75.52248781)) 

  )}}}

We see that the cosine of the indicated angle is the adjacent
side over the hypotenuse, that is

{{{cos(theta)=1/4}}}

Use a calculator to find the inverse cosine of {{{1/4}}} or {{{0.25}}}
and you get

{{{theta="75.52248781°"}}}

So each base angle has measure 75.5° (rounded to tenths). So
doubling that to account for the measures of the two base angles
we get 151°, then subtracting that from 180° gives the vertex angle
having measure 29°.

Edwin</pre>