Given - the complementary angles are equal.

Let the angle be x. So its complement  = x.
$\Rightarrow x+x={90}^{\circ }$
$\Rightarrow 2x={90}^{\circ }$
$\Rightarrow x={45}^{\circ }$
The angle which is complementary to itself is $x={45}^{\circ }$.

Steps: 1 Mark
Answer: 1 Mark
The angles of a right-angled isosceles triangle are:
${90}^{\circ }$, ${45}^{\circ }and{45}^{\circ }$.
Hence the supplementary angles are:
${180}^{\circ }-{90}^{\circ }={90}^{\circ }$
${180}^{\circ }-{45}^{\circ }={135}^{\circ }$
${180}^{\circ }-{45}^{\circ }={135}^{\circ }$

Each difference: 1 Mark
1. A line is a set of points that can be extended in either direction whereas a line segment is a part of a line, which is bounded between two given points.
2. Length of a line segment can be measured, whereas a line can extend up to infinity and cannot be measured.

Steps: 1 Mark
Result: 1 Mark
The number of endpoints in a line is zero.
$\Rightarrow X=0$
The number of endpoints in a Ray is 1
$\Rightarrow Y=1$
The number of endpoints in a line segment is 2.
$\Rightarrow Z=2$
$\therefore X+Y+Z=3$

Solution: 1 Mark each
(a) Sum of the angles should be equal to ${360}^{\circ }$
${110}^{\circ }+{60}^{\circ }+{40}^{\circ }+x^{\circ }={360}^{\circ }$
${210}^{\circ }+x^{\circ }={360}^{\circ }$
$\Rightarrow x^{\circ }={360}^{\circ }-{210}^{\circ }={150}^{\circ }$
 
(b) Let A and B be two complementary angles and let A >  ${45}^{\circ }$.
A + B =  ${90}^{\circ }$
B = ${90}^{\circ }$ − A
Therefore, B will be less than ${45}^{\circ }$.

(a) Proof: 2 Marks
(b) Solution: 1 Mark
(a)
     

We have to prove $\angle 1=\angle 3\&\angle 2=\angle 4$
$\angle 1+\angle 2={180}^{\circ }$              [Linear pair]
$\Rightarrow \angle 1={180}^{\circ }-\angle 2$  
Again, $\angle 3+\angle 2={180}^{\circ }$       [Linear pair]
$\Rightarrow \angle 3={180}^{\circ }-\angle 2$
Hence, $\angle 1=\angle 3$
Similarly, we can prove that $\angle 2=\angle 4$.
(b) $\angle$AOB = $\angle$DOE    (vertically opposite angles)
     $\therefore \angle$x = 60${}^{\circ }$

Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Sum of the angles of a triangle is ${180}^{\circ }$.
Since OC = OD,  $\angle OCD=\angle ODC={30}^{\circ }$
[Angles opposite to equal sides are equal]
$\Rightarrow \angle COD={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$
$\angle AOB=\angle COD$  [Vertically Opposite Angles]
$\therefore \angle AOB={120}^{\circ }$

(a) Steps: 1 Mark 
     Proof: 1 Mark
(b) Solution: 1 Mark
(a) $\angle XPB=\angle MSP=k$  (Alternate Angles are equal)
$\Rightarrow LX\parallel MY...\left(i\right)$.
$\angle PSM=\angle SRN=k$  (Corresponding Angles) 
 $\Rightarrow MY\parallel NZ....\left(ii\right)$.
From (i) and (ii)
$LX\parallel MYandMY\parallel NZ$
$\Rightarrow LX\parallel MY\parallel NZ$.
(b) Given  $k={50}^{\circ }$,
$\angle PSY={180}^{\circ }-k$
$={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

(a) Steps: 1 Mark
      Proof: 1 Mark
(b) Solution: 1 Mark
(a) Given, $\angle LST={105}^{\circ }$ and $\angle NTS={75}^{\circ }$
$\angle LST+\angle NTS={105}^{\circ }+{75}^{\circ }={180}^{\circ }$
These are co-interior angles.
Since co-interior angles are supplementary, lines are parallel.
Hence, the lines LM and NQ are parallel.
(b) Since AB is parallel to CD, corresponding angles will be equal.

Therefore, x = 130°

(a) Steps: 2 Marks
      Result: 1 Mark
(b) Solution: 1 Mark
(a) Since $AB\parallel CD$
$\Rightarrow \angle BAC+\angle DCA={180}^{\circ }$  
[Sum of Co-interior angles is equal to ${180}^{\circ }$]
$\Rightarrow \frac{\angle BAC}{2}+\frac{\angle DCA}{2}=\frac{{180}^{\circ }}{2}$
$\Rightarrow \angle OAC+\angle OCA={90}^{\circ }$   [AD & BC are angle bisectors]
In $\Delta AOC,$
$\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$
 [sum of  all the angles in triangle is ${180}^{\circ }$]
$\Rightarrow \angle AOC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
$\angle AOC=\angle BOD$  [Vertically Opposite Angles]
$\Rightarrow \angle BOD={90}^{\circ }$
(b)  The value of an exterior angle of a triangle is equal to the sum of the opposite interior angles.

As the exterior angle value is given to be 120${}^{\circ }$, the sum of the two interior angles is also 120${}^{\circ }$.
Thus, (x+y) = 120${}^{\circ }$