We may draw AC = 5 cm and \(\angle\)C = 30°. CA is one arm of \(\angle\)C. Point B should be lying on the other arm of \(\angle\)C. But we observe that point B cannot be located uniquely. So, the given data is not sufficient for construction of \(\triangle\)ABC.

In the given figure, \(\angle YBC=90^0\)
The angle made by first arc is 60° and second arc is 120°.
Each arc = 60°
Bisector of 60° = 30°,
So, 60° + 30° = 90°

Angle made by perpendicular bisector on a ray is equal to 90°.

Yes, RS II PQ II TU. Since, two lines parallel to a given line are also parallel to each other.

Given, line segment AB = 6 cm. If perpendicular bisector constructeg on AB. Then, the length of equal parts will be \(\frac{6}{2}=3\) cm.
∴ Perpendicular bisector divide the line in two equal parts.

In ΔLMN, we have, mㄥL = 60°, mㄥN = 120° and LM= 5 cm
Here, mㄥL + mㄥN = 60° + 120° = 180°
But the sum of two angles of a triangle should be less than 180°. So, the triangle with the given measures can not be constructed.

In ΔABC, we have,
mㄥA = 85°, mㄥB = 115° and AB = 5 cm
Here, mㄥA + mㄥB = 85° + 115° = 200° > 180°
which is not possible because the sum of all three angles of a triangle should be 180°.
So, ΔABC cannot be constructed.

Yes,we can draw ΔABC.
Here, the side AC, ㄥA and ㄥB of ΔABC are given. But to draw the triangle, we required ㄥC.
In ΔABC, by angle sum property, we have
ㄥA + ㄥB + ㄥC = 180° ⇒ 60° + 50° + ㄥC = 180°
⇒ 110° + ㄥC = 180°  ⇒ ㄥC = 180° -110° = 70°
Now, we have side AC, m ⇒ A and mㄥC.
So, we can draw MBC, by ASA criterion.