A triangle in which the difference of two sides is less than the third side can be made. The steps are as follows:
Step 1 : Draw a line segment BC of length 3.4cm.
Step 2 : Draw an angle of 45$\circ$ from point B.
Step 3 : From Ray BX cut off the line segment BD = 1.5cm.
Step 4 : Join D to C.
Step 5 : Draw bisector of DC.
Step 6 : Extend bisector of DC to intersect the Ray BX at point A.
Step 7 : Join A to C, $\Delta$ ABC is the required triangle.

In the given figure,
AP = BP (Given)
AQ = BQ (Given)
PQ is common.
So, $\Delta$ PAQ $\cong$ $\Delta$ PBQ (SSS rule).
$\Rightarrow \angle$ APQ=$\angle$ BPQ (CPCT).
Also, AP = BP (Given) and PM is common
Hence, $△$ PMA $\cong$ $△$PMB                 (SAS Rule)

So, AM = BM

As OQ is angle bisector of $\angle$POA, so $\angle$POQ = $\angle$QOA= ${45}^{\circ }$

A 60${}^{\circ }$ angle can be made by first drawing a line segment AB of any length.
Construct an angle of 60 degrees at both A and B using a compass.
Let the rays intersect each other at P. 
The triangle ABP  formed is an equilateral triangle. So we can see that the concept behind an equilateral triangle can be used in constructing a 60${}^{\circ }$ angle.

In $\Delta AXB$
PQ is the perpendicular bisector of XA, the two triangles formed are congruent by SAS rule.
$\therefore$ AB = XB
$\Rightarrow$ $\angle BAX=\angle AXB$
$\angle ABC=\angle BAX+\angle AXB$
             = $2\angle AXB=\angle LXY$
[Since $\angle AXB$ is angle bisector of $\angle LXY$]
 
$Similarly,\angle ACB=\angle MYXTherefore,\angle ABC={75}^{\circ }and\angle BCA={60}^{\circ }In\Delta ABC,wehave\angle ABC+\angle BCA+\angle CAB={180}^{\circ }{75}^{\circ }+{60}^{\circ }+\angle CAB={180}^{\circ }\angle CAB={180}^{\circ }-{135}^{\circ }\angle CAB={45}^{\circ }$

Angles 60${}^o$ and 30${}^o$ can be constructed using a scale and compass.
1. Draw a line l.
2. With O as centre and with any convenient radius, draw an arc to cut line l at A.
3. With A as centre and with the same radius, draw an arc to cut the previous arc at B.
4. Join OB and extend.  $\angle AOB={60}^o$

5. To construct 30${}^o$ we bisect 60${}^o$. 
 The remaining two angles cannot be constructed without using a protractor.