All the standard angles and the angles which can be obtained by bisecting standard angles can be constructed just by using a ruler and compass. Here except ${70}^{\circ }$ all other angles can be constructed using ruler and compass.

To construct a right angle triangle, the following steps should be followed:
1. Draw a line OB of given length.
2. With O as centre, make an arc of any radius intersecting OB at X.
3. With X as a centre, draw another arc keeping the radius same intersecting the previous arc at D.
4. $\angle$DOB will be 60${}^{\circ }$.
5. With D as a centre and keeping the radius same, draw another arc intersecting the first arc at C.
6. $\angle$COB will be 120${}^{\circ }$ .
7. Bisect COD drawing two equal arcs from each points intersecting at E.
8. Join EO and extend till A. $\angle$AOB will be 90${}^{\circ }$.
So, it can be seen that both 60${}^{\circ }$ and angle bisector construction is used in the process. 

The procedure for Triangle Construction 2 is:
1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.

2. From the ray, BX cut an arc equal to AB – AC at point P and join it to C as shown

3. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown:

4. Join AC, ∆ABC is the required triangle.

Hence, it is clear that a perpendicular bisector is required in the process.

The difference between two sides of a triangle should be smaller than the third side. So only 8 cm and 10 cm can be the possible difference.

In a triangle, the sum of the lengths of any 2 sides of a triangle must be greater than the third side. The third side can measure anything less than 11 units. Hence, the third side can be 10 units.

The information given above is insufficient to construct a triangle. So, the triangle cannot be constructed and its characteristics can’t be determined.

${75}^{\circ }=\frac{{90}^{\circ }+{60}^{\circ }}{2}$
Hence bisecting 90${}^{\circ }$ and 60${}^{\circ }$ will give 75${}^{\circ }$.

An angle bisector divides a given angle into two equal angles. Since BD divides $\angle$ ABC into two equal angles, it is an angle bisector.

It is possible to get ${30}^{\circ }$ from given angle of ${60}^{\circ }$ by simply drawing an angle bisector. Hence the given statement is true.