Question 1101510
<pre>
{{{drawing(400,700/3,-.5,5.5,-.5,3,

triangle(0,0,.6698729811,2.5,5,0),locate(4,.3,"30°"),
locate(0,0,B),locate(5,0,C), locate(.6,2.8,A),
locate(.1,1.2,c),locate(2.5,1.7,b=5m), locate(2,0,a=5m) )}}}  

Since sides a and b are both equal, &#916;ABC is isosceles,
and &#8736;A = &#8736;B.  Suppose &#8736;A = &#8736;B = x,
then since &#8736;A+&#8736;B+&#8736;C = 180°,

x + x + 30° = 180°
   2x + 30° = 180°
         2x = 150°
          x = 75°

Therefore &#8736;A and &#8736;B are 75° each.

To find the length of side c, we draw the median CD (in green),
which since &#916;ABC is isosceles, is also the perpendicular
bisector of side c and the bisector of &#8736;C.

{{{drawing(400,700/3,-.5,5.5,-.5,3,

triangle(0,0,.6698729811,2.5,5,0),locate(3.5,.23,"15°"),
locate(0,0,B),locate(5,0,C), locate(.6,2.8,A),
green(line(.3349364905,1.25,5,0)),
locate(.1,1.4,D),locate(2.5,1.7,b=5m), locate(2,0,a=5m) )}}} 

Therefore &#916;BCD is a right &#916;, and &#8736;BCD = 15°. 

We also know that BD is half of AB which is side c.

{{{BD/BC}}}{{{""=""}}}{{{opposite/hypotenuse}}}{{{""=""}}}{{{sin("15°")}}}

{{{BD/5}}}{{{""=""}}}{{{sin("15°")}}}

We multiply both sides by 5

{{{BD}}}{{{""=""}}}{{{5sin("15°")}}}

Using calculator:

{{{BD}}}{{{""=""}}}{{{1.294095226}}}

Side c is twice that or

{{{c}}}{{{""=""}}}{{{2.58819041m}}}

You can round it however you were told.

Edwin</pre>