Steps: 3 Marks
Result: 1 Mark
Given  $DE\parallel BC,\angle CBA={40}^{\circ }$
$\Rightarrow \angle ADE=\angle CBA={40}^{\circ }$   [corresponding angle]
Since, $\Delta DEF$ is equilateral
$\Rightarrow \angle DEF={60}^{\circ }$

 $\angle FEC={50}^{\circ }$   (Given)
$\angle AED={180}^{\circ }-({50}^{\circ }+{60}^{\circ })$     (Angles on a straight line)
                     = ${70}^{\circ }$
Consider $\Delta ADE$ , 
$\angle AED={70}^{\circ }$ & $\angle ADE={40}^{\circ }$ 
$\angle ADE+\angle AED+\angle DAE={180}^{\circ }$  [Angle sum property]
$\Rightarrow \angle DAE={180}^{\circ }-({70}^{\circ }+{40}^{\circ })={70}^{\circ }$.
$\therefore \angle BAC=\angle DAE={70}^{\circ }$.

(a) Solution: 2 Marks
(b) Each part: 1 Mark
(a) $\Rightarrow$ Corresponding Angles
$\angle 1=\angle 5,\angle 3=\angle 7,\angle 2=\angle 6,\angle 4=\angle 8$;
$\Rightarrow$ Vertically Opposite Angles:
 $\angle 1=\angle 4,\angle 2=\angle 3,\angle 5=\angle 8,\angle 6=\angle 7$;
$\Rightarrow$ Alternate Interior Angles:
$\angle 3=\angle 6,\angle 4=\angle 5$;
$\Rightarrow$ Alternate Exterior Angles:
$\angle 1=\angle 8,\angle 2=\angle 7$;
$\Rightarrow$ Co-interior Angles:
$\angle 4,\angle 6,\angle 3,\angle 5$.
(b) (i) $x={60}^{\circ }$ (Alternate Angles)
     (ii) $x={120}^{\circ }$ (Corresponding Angles)

(a) Steps: 1 Mark
     Result: 1 Mark
(b) Steps: 1 Mark
     Result: 1 Mark
 
(a) ${40}^{\circ }+4x+3x={180}^{\circ }$  [since POQ is a straight line]
$7x={180}^{\circ }-{40}^{\circ }$
$x=\frac{140}{7}={20}^{\circ }$
Hence, the value of 'x' is
$x={20}^{\circ }$
(b) Given, x = ${30}^{\circ }$,

So, $2x={60}^{\circ }$

Also, $2x+3y={180}^{\circ }$  (linear pair)

Replacing x in the above equation,
${60}^{\circ }+3y={180}^{\circ }$

$3y={120}^{\circ }$

$y={40}^{\circ }$

The value of $6y-3x={150}^{\circ }$

(a) Steps: 1 Mark
      Angles: 1 Mark
(b) Steps: 1 Mark
      Result: 1 Mark
(a) Given $\angle 1={45}^{\circ }$
$\Rightarrow \angle 3=\angle 1={45}^{\circ }$    [Vertically opposite angle]
$\angle 2+\angle 3={180}^{\circ }$   [Linear pair]
$\angle 2+\angle {45}^{\circ }={180}^{\circ }$   
$\angle 2={180}^{\circ }-\angle {45}^{\circ }$  
$\Rightarrow \angle 2={135}^{\circ }$
$\Rightarrow \angle 2=\angle 4$   [Vertically opposite angle]
$\Rightarrow \angle 4=\angle 2={135}^{\circ }$
(b) Let the angle be $x^{\circ }$
$\therefore$ its supplementary angle = $(180-x{)}^{\circ }$
Given that $(180-x{)}^{\circ }$ is the complementary angle of ${70}^{\circ }$
$\therefore (180-x)+70=90$
$\Rightarrow x=180-20$
$\Rightarrow x={160}^{\circ }$
The required angle is ${160}^{\circ }$.

Each part: 1 Mark
Two angles are complementary when they add up to ${90}^{\circ }$ 
Two angles are supplementary when they add up to ${180}^{\circ }$
(i) ${120}^{\circ }+{50}^{\circ }={170}^{\circ }$
Angles are neither complementary nor supplementary.
(ii) ${60}^{\circ }+{120}^{\circ }={180}^{\circ }$
Angles are supplementary.
(iii) ${39}^{\circ }+{61}^{\circ }={100}^{\circ }$
Angles are neither complementary nor supplementary.
(iv) ${65}^{\circ }+{25}^{\circ }={90}^{\circ }$
Angles are complementary.

If two angles are supplementary, then the sum of the angles will be 180${}^{\circ }$. 

Interior angle can be defined as:
When two parallel lines are crossed by another line (which is called the transversal), the pairs of angles on opposite sides of the transversal  but inside the two lines are called interior angles.
So in the given question, the interior angles are $\angle$3, $\angle$4, $\angle$5, $\angle$6.

For lines l and m, lines p and q are transversals. Similarly for lines p and q, lines l and m will be transversals.

Count the number of line segments. The number of segments required to make each letter is -

R - 5 segments

A - 3 segments

M - 4 segments

Therefore, number of line segments is 12.