Steps of construction: 
I. Draw a line segment BC = 4 cm. A,
II. At B, construct \(\angle\)CBX = 90°.
III. With centre C and radius 5 cm, draw an arc to cut \(\overrightarrow{BX}\) at A.
IV. Join AC.
Thus,  \(\triangle\) ABC is the required triangle.

Steps of construction:
I. Draw a line segment BC = 4.5 cm. 
II. Construct \(\angle\)CBX = 60° at B.
III. From BX, cut off line segment BA = 5.5 cm,
IV. Join AC
Thus, ABC is the required triangle.

Let MBC be a right triangle, right angled at C, such that hypotenuse AB = 5.4 cm. Further, let Then by the angle sum property of ΔABC, we have
\(\angle A+\angle B+\angle C\) = 180°
\(\Rightarrow\)    60° \(\angle B\) + 90° = 180°
\(\Rightarrow\)  150° + \(\angle B\) = 180°
\(\Rightarrow\)  \(\angle B\) =180° - 150° = 30°

To draw ΔABC, we follow the following steps:
(a) Draw   a line segment AB = 5.4 cm.
(b) Draw\(\angle BAX\) measure 60°.
(c) Draw \(\angle ABY\) of measure 30° with Y on the same side of AB as X.
Let AX and BY intersect at C.
Then, ΔABC is the required triangle.

Yes,we can slightly modify the construction of a line parallel to a given line through a point not on the line to use the idea of equal corresponding angles instead of equal alternate angles. For modifying this construction, we change only Step IV, VI and VII and new steps are given below:
Step IV With A as centre and radius same as in Step III draw an arc EF to cut BA (extended) at P.
Step V Place the pointed tip of the compasses at C and adjust the. opening so that the pencil tip is at D.
Step VI With the same opening as in Step V and with Pas centre, draw an arc which cut the arc EF at Q.
Step VII Join AQ to draw a line m, which is parallel to the given line I.
Here, \(\angle ABC,\angle PAQ\) are corresponding angles,

Therefore, I II m.

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

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.

PQ = 5 cm, \(\angle PQR\) =105°, \(\angle QRP\) = 40°
\(\therefore\) Sum of angles of a triangle is 180°
\(\therefore\)  \(\angle PQR+\angle QRP+\angle QPR\) = 180° 
\(\therefore\) 105° + 40° + \(\angle QPR\) = 180°
\(\therefore\) 145° + \(\angle QPR\)  = 180°
\(\angle QPR\) = 180° - 145°
\(\angle QPR\) = 35°
Steps of Construction:
(a) Draw a line segment PQ of 5 em.

(b) From point P draw an angle of 35°, such that \(\angle QPX\) = 35°
(c) From point Q draw an angle of 105°, such that \(\angle PQY\) = 150°
(d) These lines PX and QY cut at R.
PQR is required triangle.

Steps of construction
(i) Draw a line BC = 4.5 cm.
(ii) With centre B and radius 2 cm draw an arc.
(iii) With centre C and radius 4 cm draw an arc which cuts the previous arc at A.
(iv) Join AB and AC.

Hence, \(\triangle ABC\) is the required triangle in which AB= 2 cm, BC= 4.5 cm and AC= 4 cm.