Constructing Triangles

The $40^{\circ }$ angle is between the $5\text{cm}$ and $7\text{cm}$ sides, so this is Side-Angle-Side.

Draw the $7\text{cm}$ side, measure $40^{\circ }$ at one end, mark $5\text{cm}$ along the new arm and join the two ends. The third side is then forced to be $\sqrt{5^2+7^2-2\times 5\times 7\times \cos (40^{\circ })}\approx 4.51$ cm, and the other two angles are fixed as well, so exactly one triangle can be built.

The $60^{\circ }$ angle is between the two $8\text{cm}$ sides, so this is Side-Angle-Side and only one triangle can be built.

You are not told the third side - you work it out: $\sqrt{8^2+8^2-2\times 8\times 8\times \cos (60^{\circ })}=\sqrt{64}=8$ cm. So the triangle is in fact equilateral, with all three angles $60^{\circ }$, and there is only one such triangle. Flipping the drawing over gives a mirror image - the same triangle, not a second one.

Add the angles: $60^{\circ }+70^{\circ }+80^{\circ }=210^{\circ }$.

The interior angles of every triangle add to exactly $180^{\circ }$, no matter how big or small the triangle is drawn. Since $210^{\circ }$ is not $180^{\circ }$, no triangle at all can have these three angles.

The $6\text{cm}$ side lies between the $60^{\circ }$ and $50^{\circ }$ angles, so this is Angle-Side-Angle.

The third angle is forced: $180^{\circ }-60^{\circ }-50^{\circ }=70^{\circ }$. Drawing the $6\text{cm}$ base and the two arms at $60^{\circ }$ and $50^{\circ }$ from its ends, the arms cross at exactly one point, so only one triangle can be constructed. Drawing it on the other side of the base only gives a mirror image - the same triangle, not a second one.

The $10\text{cm}$ side lies between the $100^{\circ }$ and $20^{\circ }$ angles, so this is Angle-Side-Angle.

The third angle is forced: $180^{\circ }-100^{\circ }-20^{\circ }=60^{\circ }$. Drawing the $10\text{cm}$ base and the two arms at $100^{\circ }$ and $20^{\circ }$ from its ends, the arms meet at exactly one point, so only one triangle can be constructed. Flipping the drawing over gives a mirror image - the same triangle, not a second one.

The angles add to $60^{\circ }+80^{\circ }+40^{\circ }=180^{\circ }$, so triangles with these angles certainly exist.

But angles fix only the shape, not the size. Enlarging one of these triangles gives another triangle with exactly the same three angles but different side lengths, so infinitely many triangles can be drawn - they are all similar, but not congruent.