x + 2x = 180°     (Co-interior angle)
3x = 180°
x = 60°

(i) Since EF " GR and AC is a transversal.
∴ \(\angle \)CAF + \(\angle \)ACR = 180°
=> \(\angle \)CAF + 105° = 180°
=> \(\angle \)CAF = (180° - 105°) = 75°
(ii)∵ FE || GR and AC is a transversal.
:. \(\angle \)EAC = \(\angle \)ACR [Alternate angles]
=> \(\angle \)EAC = 105°
=> \(\angle \)BAC + LEAB = 105°
=> \(\angle \)BAC + 60° = 105°
=> \(\angle \)BAC = 105° - 60° = 45°
Thus, \(\angle \)CAF = 75° and \(\angle \)BAC = 45°

∵ \(\angle \)COB = 50° [Given]
∴ \(\angle \)AOD = 50° [Vertically opposite angles]
Now, \(\angle \)AOC and \(\angle \)COB form a linear pair.
Thus, \(\angle \)AOC + \(\angle \)COB = 180°
=> \(\angle \)AOC + 50° = 180°
=> \(\angle \)AOC = 180° - 50° = 130°
Also, \(\angle \)AOC and \(\angle \)BOD are vertically opposite angles.
∴ \(\angle \)BOD = \(\angle \)AOC = 130°
Thus, the required measures of the remaining
three angles are:​​​​​​​\(\begin{matrix} \angle AOD={ 50 }^{ o } \\ \angle AOC=130^{ o } \\ \angle BOD=130^{ o } \end{matrix} \)}

Points of intersection are A, B and C.
Measure of LA = 60°
Measure of LB = 60°
Measure of LC = 60°

No, two lines cannot intersect in more than one point.

Since x and 55° are vertically opposite angles
∴ x = 55°
Again, 55° + Y = 180° [Linear pair]
or y = 180° - 55°
or y = 125°
Since z and yare vertically opposite angles,
∴ z = 125° [∵ y = 125°J
Thus, x = 55°, y = 125°, Z = 125^o

(a) Yes, \(\angle \)AOB and \(\angle \)BOC are adjacent angles, because they have common vertex 0 and their outer arms (\(\overrightarrow { OA } \) and \(\overrightarrow { OC } \)) are on either side of the common arm OB.
(b) No, because \(\angle \)BOC is a part of \(\angle \)BOD.

74°

135°

Given, l || m, n and p are transversals, \(\angle \)TBO = 66°.
\(\therefore\) \(\angle \)TBO + \(\angle \)x = 180°
[\(\therefore\) l || mm and cointerior angles has a sum of 180°]
\(\Rightarrow\) \(\angle \)x =180° - 66°=114°
\(\because\) \(\angle \)OCA= \(\angle \)BDO
[alternate angles formed by transversal mare equal]
\(\because\)\(\angle \)OCA= 48°
So, \(\because\)BDO = 48°
\(\because\) \(\angle \)BDO + \(\angle \)QDO =180°     [linear pair]
\(\Rightarrow\) 48° + \(\angle \)y =180° \(\Rightarrow\) \(\angle \)y =180° - 48°=132°
So, 70° + 40° + \(\angle \)EFD = 180°
\(\Rightarrow \) 110° + \(\angle \)EFD =180°
\(\Rightarrow \) \(\angle \)EFD=180°-110° \(\Rightarrow\) \(\angle \)EFD =70°

Since, lines l₁ and l₂ are parallel to each other.
where l₃ is transversal
\(\therefore\)   \(\angle \) P and \(\angle \)130° form a linear pair angles.
So,  \(\angle \)P+130°=180°
\(\Rightarrow\) \(\angle \)P = 180° - 130°
\(\Rightarrow\) \(\angle \)P = 50°
Also, \(\angle \)P and \(\angle \)M, alternate interior angles.
\(\angle \)M=\(\angle \)P=50°

Since, angles 46° and x are vertically opposite angles.
So, x = 46°
\(\because\) Angles x and y form a linear pair angles.
\(\therefore\) x+y=180° and x = 46°
So, 46°+y=180° \(\Rightarrow\) y = 180°​​​​​​​-46°​​​​​​​
\(\Rightarrow\) y = 134°​​​​​​​
Now, angles y and z are vertically opposite angles.
​​​​​​​So, y = z = 134°​​​​​​​

Since, anglesSo,  50° and \(\left( 2x+\frac { 1 }{ 2 } x \right) \) form a linear pair angles.
So, \(50°+2x+\frac { 1 }{ 2 } x=180°\Rightarrow \frac { { 100 }^{ ° }+4x+x }{ 2 } =180°\)
\(\Rightarrow \quad 100°+5x=360°\Rightarrow 5x=360°-100\)
\(\Rightarrow \quad 5x=260°\Rightarrow \quad x=\frac { 260 }{ 5 } =52°\Rightarrow x=52°\)
So, 2x=2\(\times\)52°=104°
and \(\frac { 1 }{ 2 } x=\frac { 1 }{ 2 } \times 52°=26°\)
 

Given, p II q
\(\angle \)e + 125° = 180°  [by linear pair]
\(\Rightarrow\) \(\angle \)e = 180° -125° \(\Rightarrow\)\(\angle \)e = 55°
\(\therefore\) \(\angle \)f = \(\angle \)e = 55°      [vertically opposite angles]
Since, p II q and t is a transversal.
\(\therefore\) \(\angle \)a = \(\angle \)f = 55° [alternate interior angles]
\(\angle \)d = 125° [corresponding angles]
\(\angle \)c = \(\angle \)a = 55° [vertically opposite angles]
and \(\angle \)b = \(\angle \)d = 125° [vertically opposite angles]
Hence, \(\angle \)a = 55°, \(\angle \)b = 125°, \(\angle \)c = 55°, \(\angle \)d = 125°, \(\angle \)e = 55° and \(\angle \)f = 55°.

Let ABCD be the rectangle.
ABCD is a parallelogram with \(\angle \)A = 90°.
We have, \(\angle \)C = \(\angle \)A = 90°
[opposite angles of a parallelogram are equal]

Again, \(\angle \)A + \(\angle \)B= 180°
[\(\because\) \(\angle \)A and \(\angle \)B are adjacent angles of a parallelogram]
\(\Rightarrow\) 90° + \(\angle \)B= 180° \(\Rightarrow\)  \(\angle \)B= 180° - 90° = 90°
\(\therefore\) \(\angle \)D = \(\angle \)B= 90°
[\(\because\) opposite angles of a parallelogram are equal]
Hence, \(\angle \)A = \(\angle \)B = \(\angle \)C = \(\angle \)D = 90°