Steps of Construction:
1. Draw a line segment BC of length 5 cm.
2. At B, draw LMBC = 60° and produced line BM.
3. From point C draw a line making an angle of 30°.
4. Both the lines intersect at A.
5.MBC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Locate three points B1, Bz, and B3on line segment BX.
8. Join BC
9. Draw a parallel line through B3to B3C intersecting extended line BCat C.
10. Through C' draw a line parallel to AC intersecting extended line segment BA at A'. A'BC is the required triangle.

Steps of construction:
1. Draw a line segment BC = 8 cm.
2. Construct AM..L BC
3. Taking C as centre and radius as 10 cm, draw an arc that s the ray BMat A'.
4. Join CA' to obtain ~ ABC A
5. Below BC, make an acute angle CBX.
6. Along BX mark off S points B₁,B₂,B₃,B₄,B₅ such that BB₁ = B₁B₂= B₂B₃= = B₄B₅.
7. Join BC
8. From B₄'Draw B₄C' II BC
9. From aC' draw CA' II CA meeting BA at point A'. Then A' BC' is the required triangle.

Steps of construction:
1. Draw a circle of radius 4 cm with 0 as centre.
2. Take a point A on the circumference of the circle and join OA. Draw perpendicular to OA at point A.
3. Draw a radius OB, making an angle of 1200 with OA.
4. Draw the perpendicular to OB at point B. Let both the perpendiculars intersect at point P.
5. Join OP. PA and PB are required tangents, which make an angle of 600 to each other.

Given A\(\Delta ABC\) , in which BC=5cm, \(\angle CAB=120°\) and \(\angle ABC=30°\) .
Then, \(\angle BCA=180°-30°-120°=30°\)

No, two triangles are not congruent.

Given A right-angled triangle with sides of lengths 8 cm and 6 cm making a right angle. Required Triangle whose sides are \(\frac { 3 }{ 4 } \) times of the corresponding sides of given triangle.
Steps of Construction
1. Construct a right angled \(\angle ABC\)   right angle at B with sides BC = 8 cm and AB = 6 cm
2. Through B, construct an acute LCBX on the side \(\angle CBX\) opposite to the vertex A.
 
3. Mark four points B₁, B₂ , B₃, and B₄ on BX such that BB₁ = B₁B₂ = B₂B₃,= B₃B₄
4. Join B₄C
5. Through B₃, draw B₃C' II B₄C intersecting BC .at C'.
6. Through C', draw C'A' II CA intersecting AB at A'. Hence, \(\Delta A'BC'\)  is the required triangle.

Given A \(\Delta ABC\)  in which BC = 7 cm, \(\angle B=45°\) and \(\angle C=60°\) Required Draw \(\Delta A'BC'-\Delta ABC\)with scale factor \(\frac { 3 }{ 5 } <1\)
Steps of Construction
1. Draw a line segment BC = 7 cm.
2. At B, construct an \(\angle CBY=45°\)
3. At C, construct an \(\angle BCA=60°\)  intersecting BY at A.
     
4. Join AB and CA. Then, \(\Delta ABC\) is the required triangle. 
5. Through B, construct an acute angle \(\angle CBX\)  on the side opposite to the vertex A.
6. Mark five points B₁, B₂, B₃, B₄ and B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ =B₃B₄ = B₄B₅
7. Join B₅C

yes

In \(\Delta\)PQR 
\(\angle\)Q = \(\angle\)B, PQ = PR = 8 cm.
\(\therefore\)      \(\angle\)R = \(\angle\)Q = \(\angle\)B
\(\therefore\)    Also    \(\angle\)B = \(\angle\)C
\(\Rightarrow\)   \(\angle\)A=\(\angle\)P
In \(\Delta\)ABC and \(\Delta\)PQR
\(\frac { AB }{ AC } =\frac { PQ }{ QR } \) , \(\angle\) A = \(\angle\)P
\(\therefore\)  \(\Delta\)ABC \(\sim \) \(\Delta\) PQR

\(\frac { AP }{ AB } =\frac { 3 }{ 4 } ;\frac { AQ }{ AC } =\frac { 1 }{ 4 }\)
\( \\ PQ\cong 3.25\quad cm\)