Question 444392
<pre>
{{{drawing(400,400,-4,4,-4,4,

arc(0,0,6,-6,90,180), triangle(0,3,1.5sqrt(3),1.5,0,0),
green(arc(0,0,1,-1,240,390)),red(locate(.35,-.35,x)),


triangle(-3,0,-1.5,-1.5sqrt(3),0,0) )}}} 

All the interior angles of all equilateral triangles 
have measure 60°

The angle in the quarter circle is 90°.

{{{drawing(400,400,-4,4,-4,4,

arc(0,0,6,-6,90,180), triangle(0,3,1.5sqrt(3),1.5,0,0),
green(arc(0,0,1,-1,240,390)),red(locate(.35,-.35,x)),
locate(.1,.6,"60°"), locate(-.7,-.1,"60°"),

locate(-.5,.4,"90°"),

triangle(-3,0,-1.5,-1.5sqrt(3),0,0) )}}} 
 
So the sum of the measures of all the marked angles
is 360°, so we have the equation:

     60° + 90° + 60° + x° = 360°

                210° + x° = 360°
 
                       x° = 150°

Edwin</pre>