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 QR = 5 cm.
II. At Q, construct \(\angle\)RQX = 60°.
III. At R, construct \(\angle\)QRY = 75°
IV. Let the rays \(\overrightarrow{QX}\) and \(\overrightarrow{RY}\) intersect at P.
Thus \(\triangle\)PQR is the required triangle.

Steps of Construction:
Step I: Firstly, we draw a rough sketch of triangle with given measures marked on it.

Step I: Draw a line segment AB of length 4.5 cm,
Step II: Draw an angle of 90° on point Band produce it to Y.
Step III: With B as centre, draw an arc of 4.5 cm which intersects ray BY to C
Step IV: Join CA.

Thus, ΔABC is the required right angled isosceles triangle.

In order to draw the ΔABC and the perpendicular bisector of BC, we follow the following steps:
(a) Draw a line segment BC = 5 cm,
(b) With centre B and radius AB = 4.5 cm, draw an arc of the circle

(c) With centre C and radius AC = 7 cm, draw an arc intersecting the previously drawn arc at A.
(d) Join AB and AC to obtain the desired triangle.
(e) With centre B and radius more than \(\frac { 1 }{ 2 } \) (BC), draw two arcs on both sides of BC.
(f) With centre C and the same radius as in step (v) draw two arcs intersecting the arcs drawn in step (v) at D and E.
(g) Join DE to obtain the required perpendicular bisector of BC.

To construct the MBC, we follow the following
steps:
(a) Draw a line segment BC = 7 cm,
(b) With centre Band radius AB = 6 cm, draw an arc of the circle.
(c) With centre C and radius AC = 5 cm, draw another arc intersecting the arc drawn in step (iii) at A.
(d) Join AB and AC to obtain the desired triangle.

Here, we are given the side BC, \(\angle B\) and \(\angle A.\) But to draw the triangle, we required \(\angle C.\)
We know that
\(\angle A+\angle B+\angle C\) = 180°
\(\Rightarrow\) 100° + 60° \(\angle C\) = 180°
\(\Rightarrow\) 160° + \(\angle C\) = 180°
\(\Rightarrow\)    \(\angle C\) = 180° - 160° = 20°
Thus, we have Bc, = 5.2 cm, \(\angle B\) = 60° and \(\angle C\) = 20°. Now, to draw the Δ ABC, we follow the following steps:
(a) Draw a line segment BC = 5.2 cm.
(b) Draw \(\angle CBX,\) such that \(\angle CBX\) = 60°

(c) Draw \(\angle BCY,\) with Y on the same side of BC as X such that \(\angle BCY\) = 20°
Let BX and CY intersect at A.
Then ΔABC 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.

To construct the MBC, we follow the following steps:
(a) Draw a line segment BC of length 4.5 cm 

(b) Draw
(c) With centre Band radius AB = 6.2 cm, draw an arc of the circle to intersect ray CX at A.
(d) Join BA to obtain the desired triangle ABC.

Steps of Construction:
Part I: Construction of ΔABC :
(a) Draw BC = 5.4 cm.
(b) Construct
(c) From BX cut off BA = 4.5 cm,
(d) Join A and C.
Now ΔABC is the required triangle.
Part II: Construction of AD丄BC:
(a) Produce CB through B to Y.
(b) With centre A and a sufficient radius, draw an arc intersecting BY and BC at U and M respectively.
(c) With U as centre and a radius more than half of UM, draw an arc opposite to side A.
(d) With the same radius and centre M, draw another arc, cutting the previous arc at T.
(e) Joint AT such that it meets YB at D. Then AD丄 Be.

Steps of Construction:
(a) Draw a line segment PQ = 5 cm,
(b) At P, construct
(c) At Q construct (d) Let the rays \(\overline { PX } \) and \(\overline { QY } \) intersect at R.
Thus, ΔPQR is the required triangle.

∠ABC = 90°
So, ΔABC is a right angled triangle.

From the figure, we get \(\angle ACB=\angle ABC=60^0\)
\(\therefore\triangle ABC\) is an isosceles triangle \(\Rightarrow\) AC = AB = 6 cm

In ΔABC, we have mㄥA = 70°, mㄥB = 50° and AC=3 cm.
Here, two angles and one side AC are given.
So, to construct the triangle, we need mㄥC
By angle sum property of a triangle,
mㄥA + mㄥB + mㄥC = 180°
70° + 50° + mㄥC = 180° ⇒ 120° + mㄥC = 180°
⇒ mㄥC = 180° -120° = 60°

No, we cannot construct a ΔDEF such that EF = 7.2 cm, mㄥE = 110° and mㄥF = 80°
Justification
We know that, the sum of all the three angles of a triangle is 180°. But in given question, sum of two angles.
mㄥE + mㄥF
= 110° + 80° = 190° > 180°
The sum of these two angles should be less than 180°.
So, the triangle with given measures cannot be constructed.