Maths M.C.Q. [1 Marks Each]

 Answer is $(B)$
If two adjacent angles measures $90$ degrees only when lines are perpendicular to each other.
hence the above statement is False.

 Let the angle be $X$ It is given that $X$ is its own complementary angle.
$\Rightarrow X + X = 90^\circ $
$\Rightarrow 2X = 90^\circ $
$\Rightarrow X = 45^\circ $

 Supplementary angles are two angles that have a sum of $180^\circ .$
The given supplementary angle is $100^\circ $ and we have to find other.
$\Rightarrow $ Supplementary angle $= 180^\circ - 100^\circ = 80^\circ .$
$\therefore$ Supplementary angle of $100^\circ $ is $80^\circ .$

The pair of angles is said to be complementary, when their sum is $90^\circ .$
Let $x, y$ be any two complementary angles and $\text{m}(\angle{\text{x}})=90^\circ$
$\Rightarrow\text{m}(\angle{\text{x}})+\text{m}(\angle{\text{y}})=90^\circ.....$ (By definition of complementary angles)
$\Rightarrow90^\circ+\text{m}(\angle\text{y})=90^\circ$
$\Rightarrow\text{m}(\angle{\text{y}})=0^\circ$
Hence, measure of complementary angle of $90^\circ $ is $0^\circ .$

Let the complementary angles be $x$ and $(90^\circ − x).$
Then, $2x = 3(90^\circ − x)$
$\Rightarrow2{\text{x}}=270^\circ-3\text{x}$
$\Rightarrow5\text{x}=270^\circ$
$\Rightarrow\text{x}=\frac{270^\circ}{5}=54^\circ$
$\therefore$ The two angles are $54^\circ , 36^\circ $

 Let the required angle be $x,$ then its supplement $= (180 - x)$ and $x = 2(180 - x)$
So, $(180 - x) + 2(180 - x) = 180$
$\Rightarrow 180 - x + 360 - 2x = 180$
$\Rightarrow -3x + 360 = 0$
$\Rightarrow -3x = -360$
$\Rightarrow 3x = 360$
$\Rightarrow x = 120^\circ $

Sum of the given angles $= x + 90^\circ – x = 90^\circ $
Since, the sum of given two angles is $90^\circ $
Hence, they are complementary to each other.

Let the desired angle be $x.$
According to the question,
$\Rightarrow\text{x}=\frac{1}{5}(180-\text{x})$
$\Rightarrow6\text{x}=180$
$\Rightarrow\text{x}=30^\circ$

Let the required angle be $x$
Now, supplementary of the required angle $= 180^\circ- x$
Then,
$\text{x}+\frac{1}{3}(180^\circ-\text{x})=90^\circ$
$\Rightarrow 3\text{x}+180^\circ-\text{x}=270^\circ$
$\Rightarrow 2\text{x}=90^\circ$
$\Rightarrow \text{x}=45^\circ$
Hence, the correct answer is option $(d).$

Since, $POR$ is a line. So, the sum of angles forming linear pair is $180^\circ .$
$\therefore\angle\text{POQ}=\angle\text{ROQ}=180^\circ$
$\Rightarrow(3\text{a}+5)^\circ+(2\text{a}-25^\circ)=180^\circ$
$\Rightarrow3\text{a}+5^\circ+2\text{a}-25^\circ=180^\circ$
$\Rightarrow5\text{a}-20^\circ=180^\circ$
$\Rightarrow5\text{a}=180^\circ+20^\circ$
$\Rightarrow5\text{a}=200^\circ$
$\Rightarrow\text{a}=\frac{200^\circ}{5}$
$\Rightarrow\text{a}=40^\circ$
Hence, the value of a is $40^\circ .$

Let one angle be $x^\circ $
then other angle is $2x^\circ ($As twice of one angle$)$
Since two angles are supplementary
$x^\circ + 2x^\circ = 180^\circ $
$3x^\circ = 180^\circ $
$\text{x}=\frac{180}{3}=60^\circ$
One angle is $60^\circ ,$
Other angle $2x^\circ = 2 \times 60^\circ = 120^\circ $

Linear pair of angles are supupsupplementary ie they add up to form $180$ degrees.
Hence one angle of linear pair is acute, other has to be obtuse angle.

 Two angles which are supplementary add upto $180 .....(1)$
Let one angle be $x$ and other be $\frac{1}{5}\text{x}$
Hence, $\text{x}+\frac{1}{5}\text{x}=180....$ From $(1)$
$\Rightarrow\frac{6}{5}\text{x}=180$
$\Rightarrow\text{x}=180\times\frac{5}{6}=150$
Thus one angle is $150$ and the other angle is
$\frac{150}{5}=30^\circ$

$\angle \text{AOC}$ and $\angle \text{BOC}=180^\circ$ [$\because$ Linear pair angles]
$\Rightarrow 44^\circ+(2\text{x}+6)^\circ=180^\circ$
$\Rightarrow (2\text{x+6})^\circ=136^\circ$
$\Rightarrow 2\text{x}+6=136$
$\Rightarrow 2\text{x}=130$
$\Rightarrow \text{x}=65$
Hence, the correct answer is option $(b).$

Let the angle be $\angle{\text{x}}$
Thus, according to the question,
$x = 14 + 90 - x$
$\Rightarrow 2x = 104$
$\Rightarrow x = 52$
Therefore, the angle is $52^\circ .$

Complementary angles are those, whose measures add up to $90^\circ $ Out of the given options,
$30+150=180^\circ\neq90^\circ76+14=90^\circ\\65+65=130^\circ\neq90^\circ120+30=150^\circ\neq90^\circ$

$\angle\text{POR}=3\angle\text{ROQ}$ (Given)
$\angle\text{POR}+\angle\text{ROQ}=180^\circ$
$\Rightarrow3\angle\text{ROQ}+\angle\text{ROQ}=180^\circ$
$\Rightarrow4\angle\text{ROQ}=180^\circ$
$\Rightarrow\text{ROQ}=\frac{180^\circ}{4}=45^\circ$
Now, $\angle\text{SOQ}=\angle\text{POR}$
$=3\angle\text{ROQ }(\text{Ver. opp.}{\angle\text{S}})$
$=3\times45^\circ=135^\circ$

Let the angles be $2x$ and $7x$
Angles are given supplementary $2x + 7x = 180^\circ $
$9x = 180^\circ $
$x = 20^\circ $
So the angles are $2 \times 20^\circ = 40^\circ $ and $7 \times 20^\circ = 140^\circ $

Let unknown angle be $x^\circ .$
$\therefore$ Complement of $x = 90^\circ - x$
Acc to question,
$(90 - x)^\circ - x = 56^\circ $
$90^\circ - 2x = 56^\circ $
$-2x = -34^\circ $
$x = 17^\circ $
$\therefore (90 - x) = 90^\circ - 17^\circ = 73^\circ $

A Line $AB$ is parallel to the line $CD.$
This is symbolically written as
$\overleftrightarrow{\text{AB}}\parallel\overleftrightarrow{\text{CD}}$

Let the other angle be $x$
Now their sum $= 100^\circ $
$\Rightarrow x + 35^\circ = 100^\circ $
$\Rightarrow x = 100^\circ - 35^\circ = 65^\circ $
So the other angle is $65^\circ $

From the above figure, we can say that angle between South and West is $90^\circ $ and angle between South and East is $90^\circ .$ So, their sum is $180^\circ .$
Hence, both angles make a linear pair.

Since, $\angle\text{POR}, \angle\text{ROT}$ and $\angle\text{TOQ}$ lies on a straight line $POQ,$
then their sum is equal to $180^\circ .$
$\therefore\angle\text{POR}+\angle\text{ROT}+\angle\text{TOQ}=180^\circ$
$\Rightarrow 90^\circ + x + y = 180^\circ $
$\Rightarrow x + y = 180^\circ - 90^\circ $
$\Rightarrow x + y =90^\circ ...(i)$
Also, $x : y = 3 : 2 [$given$]$
Let, $x = 3a$ and $y = 2a$
$\therefore 3a + 2a = 90^\circ [$from Eq$.(i)]$
$\Rightarrow 5a = 90^\circ $
$\Rightarrow \text{a}=\frac{90^\circ}{5}=18^\circ$
Now, $x = 3a = 3 \times 18^\circ = 54^\circ $ and $y = 2a = 2 \times 18^\circ = 36^\circ $
Since, y and z from a liner pair.$\therefore y + z 180^\circ $
$\Rightarrow 36^\circ + \text{z}=180^\circ  \Rightarrow \text{z}=180^\circ-36^\circ [\because\text{y}=36^\circ]$
$\Rightarrow \text{z}=144^\circ$

 Two angles are Complementary when theyadd up to $90$ degrees.
So, third angle will be $90.$
So, it will be right angle triangle.So Punita wants to classify
$(A)$ Right, because complementary angles add up to $90$ and the difference between $180$ and $90$ is $90.$

 $\angle1+\angle2=180^\circ$ (Linear pair)
$\Rightarrow5\text{x}+15^\circ+28\text{x}=180^\circ$
$\Rightarrow33\text{x}=180^\circ-15=165^\circ$
$\Rightarrow\text{x}=\frac{165^\circ}{33}=5^\circ$

Since, $AB || CD$
$\therefore \angle \text{BPQ}=\angle \text{DQF}$ [Corresponding angles]
$\Rightarrow (5\text{x}-20)^\circ=(3\text{x}+40)^\circ$
$\Rightarrow 5\text{x}-20=3\text{x}+40$
$\Rightarrow 2\text{x}=60$
$\Rightarrow \text{x}=30$
$\therefore \text{BPQ}=(5\times 30-20)^\circ=130^\circ$
Now, $\angle \text{APE}=\angle \text{BPQ}$ [Vertically opposite angles]
$\Rightarrow 2\text{y}^\circ=130^\circ$
$\Rightarrow \text{y}=65$
$\therefore \text{y}-\text{x}=65-30$
$=35$
Hence, the correct answer is option $(b).$

 Two angles are complementary when they add upto form $90$ degrees.
If one angle $= x$
complemen of angle $= 90 - x$
Hence $x = 90 - x -28 ($given$)$
$2x = 90 - 28$
$2x = 62$
$\text{x}=\frac{62}{2}=31$
$= 31$
Hence angle is $31$ and its complementis $59$

Let the complement be $y$ Two angles are complementary if their sum is $90^\circ$
$\therefore 90^\circ - a + y = 90^\circ$
$⇒ y = a$

 $\angle\text{A}=\angle\text{B }(\text{Vertically opp. angles})$
$\Rightarrow 11n - 13 = 7n + 39^\circ $
$\Rightarrow 11n - 7n = 39^\circ + 13^\circ $
$\Rightarrow 4n = 52^\circ $
$\Rightarrow n = 13^\circ $

 Let the required angle be $x$
Complementary angles $=$ Sum of two angles is $90^\circ $
$\therefore x = (90 - x) + 20$
$x = 90 - x + 202x = 110$
$\therefore x = 55^\circ $

$ (8x - 41)^\circ + (3x)^\circ + (3x + 10)^\circ + (4x - 5)^\circ = 360^\circ $
$\Rightarrow 8x - 41 + 3x + 3x + 10 + 4x - 5 = 360$
$\Rightarrow 18x - 36 = 360$
$\Rightarrow 18x = 396$
$\Rightarrow x = 22$
Hence, the correct answer is option $(a).$

Let the required angle be $x,$ then its complement $= (90 - x)$
Given that $x = (90 - x) + 30$
$\Rightarrow 2x = 120$
$\Rightarrow x = 60^\circ $

 Let the angles be $3x$ and $7x.$
We know that sum of two interior angles on the same side of transversal is $180.$
$3x + 7x =180$
$\Rightarrow 10x = 180$
$\Rightarrow x = 18$
Therefore, the greater angle is $7x = 7 \times 18 = 126$

Let the angle in degrees be $x$. Then, its supplement $= (180^\circ − x)$
Given, $\text{x}=\frac{1}{5}(180^\circ-\text{x})$
$\Rightarrow 5x = 180^\circ − x$
$\Rightarrow 6x = 180^\circ $
$\Rightarrow x = 30^\circ $

 $\angle$ between the lines $x + y - 3 = 0\ \&\ x - y + 3 = 0$ is $90^\circ $
$\Rightarrow\alpha=90^\circ$
As, $\beta$ is acuteTherefore $\alpha>\beta$

 Given, $PQ || SR$ and $PR$ is transversal.
$\therefore \angle \text{QPR}=\text{SRP}$ [Alternate interior angles]
$\Rightarrow \text{a}=20^\circ$
Also , $SP || RQ$ and $PR$ is transversal.
$\therefore \angle \text{SPR}=\text{QRP}$ [Alternate interior angles]
$\Rightarrow \text{b}=50^\circ$

 Let the $\angle{\text{x}}$ be the required angle
Thus, according to the question
$180 - x = 3 (90 - x)$
$\Rightarrow 180 - x = 270 - 3x$
$\Rightarrow 2x = 90$
$\Rightarrow x = 45$

$\angle \text{AOD}+\angle \text{BOD}=180^\circ$ [Linear pair angles]
$\Rightarrow (7\text{x}-20)^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 7\text{x}-20+3\text{x}=180$
$\Rightarrow 10\text{x}=200$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{AOD}=(7\times 20-20)^\circ=120^\circ$
Now, $\angle \text{AOD}=\angle \text{BOC}=120^\circ$ [Vertically opposite angles]
$\therefore \text{y}=120$
Now,$ x + y = 20 + 120$
$= 140$
Hence, the correct answer is option $(b).$

Let the required angle be $x$
Now, supplementary of the required angle = $180^\circ$$- x$
Then,
$x = 2$($180^\circ$ $- x$)
$⇒ x =$ $360^\circ$ $- 2x$
$⇒ 3x =$ $360^\circ$
$⇒ x = $$120^\circ$
Hence, the correct answer is option $(b).$

 Given that the ratio of two complementary angles is $13 : 5.$
Let the angles are $13x$ and $5x$ We know, $13x + 5x = 90^\circ $
$\Rightarrow x = 5$
Therefore, the angles are $65^\circ $ and $25^\circ .$

 $\angle \text{BQF}=\angle \text{AQP}=(4\text{x})^\circ$ [Vertically opposite angles]
Since, $AB || CD$
$\therefore \angle \text{AQP}+ \angle \text{CPQ}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow (4\text{x})^\circ+(5\text{x})^\circ=180^\circ$
$\Rightarrow 9\text{x}=180$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{BQF}=(4\times20)^\circ=80^\circ$
Now, $\angle \text{BQF}=\angle \text{DPQ}=80^\circ$ [Corresponding angles]
Hence, the correct answer is option $(b).$

 $\angle \text{AOD}+\angle \text{DOB}+\angle \text{BOC}=180^\circ [ \because AOC$ is a straight line$]$
$\Rightarrow 38^\circ+\text{x}+90^\circ=180^\circ$
$\Rightarrow \text{x}+128^\circ=180^\circ$
$\Rightarrow \text{x}=52^\circ$
Hence, the correct answer is option $(b).$

 If the angle is $x^\circ ,$ then by hypothesis
$\Rightarrow x^\circ = 8 (90^\circ - x^\circ )$
$\Rightarrow 720 = 8x^\circ + x^\circ $
or $ 9x^\circ = 720^\circ $
$\therefore x = 80^\circ $

 Let the angle be $x$
$\therefore$ complement $= (90 - x) x + (x + 20)$
$= 902x = 90 - 20$
$\text{x}=\frac{70}{2}$
$x = 35^\circ $
$\therefore$ other angle $= 90 - 35 = 55^\circ $

 Let all the lines intersect at $O.$

$\angle \text{COF}=\angle \text{DOE}=4\text{x}^\circ$ [Vertically opposite angles]
$\angle \text{AOC}+\angle \text{COF}+\angle \text{BOF}=180^\circ [AOB$ is a straight line$]$
 $\Rightarrow 2\text{x}^\circ+4\text{x}^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 9\text{x}^\circ=180^\circ$
$\Rightarrow9\text{x}=180$
$\Rightarrow \text{x}=20$
Hence, the correct answer is option $(c).$

Sum of supplementary angles $= 180^\circ $
Given two supplementary angles are in the ratio $5 : 7.$
$1^\text{st} \text{angle}=\frac{5}{12}\times180^\circ=5\times15=75^\circ$
$2^\text{st} \text{angles}=\frac{7}{12}\times180^\circ=7\times15=105^\circ$

Since, $AB || CD$
$\therefore \angle \text{BPQ}= \angle \text{PQC}$ [Alternate interior angles]
$\Rightarrow(3\text{x}+34)^\circ=(5\text{x}-14)^\circ$
$\Rightarrow 3\text{x}+34=5\text{x}-14$
$\Rightarrow 48=2\text{x}$
$\Rightarrow \text{x}=24$
$\therefore \angle\text{BPQ}=(3\times 24+34)^\circ=106^\circ$
$\angle \text{BPQ}+\angle \text{BPE}=180^\circ$ [EF is a straight line]
$\Rightarrow 106^\circ+\angle \text{BPE}=180^\circ$
$\Rightarrow \angle \text{BPE}=74^\circ$
Hence, the correct answer is option $(c).$

 If two angles are supplementary, then the sum of the angles is $180^\circ .$
If the ratio is $4 : 5,$ let angles are $4x$ and $5x$
Now we know, $4x + 5x = 180^\circ $
$\Rightarrow 9x = 180$
$\Rightarrow x = 20$
Therefore, angles are $100^\circ $ and $80^\circ $

 For the lines $mm$ and $nn$ to be parallel corresponding angles should be equal, i.e,
$\angle1=\angle5$
$\Rightarrow26\text{x}-7^\circ=20\text{x}+17^\circ$
$\Rightarrow6\text{x}=24^\circ$
$\Rightarrow\text{x}=4^\circ$

Two supplementary angles differ by $44^\circ $
$\therefore x + (x + 44^\circ ) = 180^\circ $
$2x = 136^\circ $
$x = 68^\circ $
Other angle $= (x + 44^\circ ) = (68^\circ + 44^\circ ) = 112^\circ $

 We have, $a : b = 3 : 2$ Let $a = 3x$ and $b = 2x.$
Since, $a$ and $b$ form a linear pair.
$\because \text{a}+\text{b} = 180^\circ$
$\Rightarrow 3\text{x} + 2\text{x} = 180^\circ$
$\Rightarrow 5\text{x} = 180^\circ [ \because$ sum of linear of angles is $180^\circ ]$
$\Rightarrow\text{x}=\frac{180°}{5}$
$\Rightarrow \text{x} = 36^\circ $
$\therefore \text {a}=3\text{x}\Rightarrow\text{a}=3\times36^\circ=108^\circ$
Now, $f = a [$Corresponding angles$]$
$\Rightarrow \text{f} = 108^\circ$

 Supplementary angles are those whose sum is $180$ degrees.
Complementary angles are those angles whose sum is $90$ degrees.
hence they dont need to be adjacent.

$\text{x}=\frac{\text{x}-1}{\text{x}+1}$
Slope of tangent at
$\text{(x, y)}=\frac{\text{dy}}{\text{dx}}=\frac{2}{(\text{x}+1)^2}$
for tangent to be parallel to $y = 2x + 1$
$\frac{2}{(\text{x}+1)^2}=2$
$\Rightarrow{(\text{x}+1)}^2=1$
$\Rightarrow\text{x}=0$ or $\text{x}=-2$
$\Rightarrow $ corresponding points are $(0, -1)$ & amp; $(-2, 3)$

From them given figure, $\angle1=\angle5$ [Corresponding angle]
$\Rightarrow\angle5 = (2\text{a}+\text{b})^\circ$
$[\because\angle1=(2\text{a}+\text{b})^\circ,\text{given}]$
Also, $\angle5+\angle6=180^\circ$ [liner paire]
$\Rightarrow (2\text{a} + \text{b})^\circ + (3\text{a} - \text{b})^\circ = 180^\circ$
$[\because\angle6=(3\text{a}-\text{b})^\circ,\text{given}]$
$\Rightarrow(2\text{a} + 3\text{a}) + (\text{b} - \text{b}) = 180^\circ$
$\Rightarrow5\text{a} = 180^\circ$
$\Rightarrow\text{a}=\frac{180^\circ}{5}$
$\Rightarrow \text{a} = 36^\circ$
Now, $\angle1+\angle2=180^\circ$ [liner paier]
$\Rightarrow\angle2=180^\circ-\angle1$
$\Rightarrow\angle2=180^\circ-(2\text{a}+\text{b})^\circ$
$[\because\angle1=(2\text{a}+\text{b})^\circ,\text{given}]$
$\Rightarrow\angle2=180^\circ-2\text{a}-\text{b}$
$\Rightarrow\angle2=180^\circ-2\times36^\circ-\text{b}$
$[\because\ \text{a}=36^\circ]$
$\Rightarrow\angle2=180^\circ-72^\circ-\text{b}$
$\Rightarrow\angle2=(180^\circ-\text{b})^\circ$

 
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
 Since, $PQ || AB$
$\therefore \angle \text{AME}+\angle \text{QEM}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 139^\circ+\angle \text{QEM}=180^\circ$
$\Rightarrow \angle \text{QEM}=41^\circ$
Now, $\angle \text{QEM}+\angle \text{DEQ}=\angle \text{MED}$
$\Rightarrow 41^\circ+\angle \text{DEQ}=67^\circ$
$\Rightarrow \angle \text{DEQ}=26^\circ$
Now, $\angle \text{PED}+\angle \text{DEQ}=180^\circ$ [Linear Pair angles]
$\Rightarrow \angle \text{PED}+26^\circ=180^\circ$
$\Rightarrow \angle \text{PED}=154^\circ$
Since, $ PQ || AB$
$\therefore \text{x}^\circ=\angle \text{PED}$ [Corresponding angles]
$\Rightarrow \text{x}^\circ-154^\circ$
$\Rightarrow \text{x}=154$
Hence, the correct answer is option $(a).$

Lets say an angle of a triangle is $\theta ,$ then after it is bisected each smaller angle will now be $\frac{\theta}{2}$​.
If you take the external angle of $\theta ,$ it will be $180^\circ - \theta $ and if this is bisected the two new angles would be
$\frac{(180^\circ-\theta)}{2}$ which is equal to $\frac{90^\circ-\theta}{2}$ .
Now adding these two angles, we get $\frac{\theta}{2}+\frac{90^\circ-\theta}{2}=90^\circ.$
Hence the angle between the internal and external bisectors of an angle of a triangle is always $90^\circ .$

Let $x, y$ be any two supplementary angles and $x$ be the smaller angle.
$\therefore x + y = 180^\circ $
Also, $x = 4 (360^\circ - x)$
$⟹ x = 4 \times 360^\circ - 4x$
$⟹ 5x = 4 \times 360^\circ $
$⟹ x = 4 \times 72 = 288^\circ x + y = 180^\circ $
$⟹ y = -108^\circ = 252y = -108^\circ = 252^\circ $
$\therefore x − y = 288 - 252 = 36^\circ $

Let two angles be $x$ and $(180 - x)$
According to question,
$x = 4(90 − x)$
$\Rightarrow x = 360 − 4x$
$\Rightarrow 5x = 360$
$\Rightarrow x = 72^\circ $
$\therefore$ Angles are $72^\circ $ and $108^\circ $
Difference of these two angles $= 108^\circ - 72^\circ = 36^\circ .$

 Two angles are called to be supplementary if the summation of both angles is $180^\circ $
say$\angle{\text{A}}$ and $\angle{\text{B}} $ are supplementary angles
$\Rightarrow\angle{\text{A}}+\angle{\text{B}}=180^\circ$
$\therefore\angle{\text{A}}=4\times\angle{\text{B}}$ (Given in question)
So, $4\times\angle{\text{B}}+\angle{\text{B}}=180^\circ$
$\Rightarrow\angle{\text{B}}=36^\circ$
Now, $\angle{\text{A}}=4\times\angle{\text{B}}$
$\Rightarrow\angle{\text{A}}=4\times36^\circ$
$\Rightarrow\angle{\text{A}}=144^\circ$
so these angles are $36^\circ , 144^\circ $

 It is given that, $PA || BC$ and $AB$ is transversal.
$\therefore\angle\text{PAB}=\angle\text{ABC}$ [Alternate interior angles]
$\Rightarrow 50^\circ=\text{a}$
Also, $AB || DC$ and $BC$ is transversal.
$\therefore\angle\text{ABC}+\angle\text{DCB}=180^\circ$ [Consecutive interior angles]
$\Rightarrow\text{a}+\angle\text{DCB}=180^\circ$
$\Rightarrow\angle\text{DCB}=180^\circ-\text{a}$
$\Rightarrow\angle\text{DCB}=180^\circ-50^\circ$ $[\because\text{a}=50^\circ]$
$\Rightarrow\angle\text{DCB}=130^\circ$
Also, $BC || DT$ and $DC$ is transversal.
$\therefore\angle \text{CDT}=\text{DCB}$ [Alternate interior angles]
$\Rightarrow \text{b}= 130^\circ$ $[\because\angle\text{DCB}=130^\circ]$

Given, $2\angle\text{A}=3\angle\text{B}=6\angle\text{C},$
$2\angle\text{A}=6\angle\text{C}\angle\text{A}=3\angle\text{C}$
$3\angle\text{B}=6\angle\text{CB}=2\angle\text{C}$
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$3\angle\text{C}+2\angle\text{C}+\angle\text{C}=180^\circ$
$6\angle\text{C}=180^\circ$
$\angle\text{C}=30^\circ$
Small angle $= 30^\circ $

 Let one angle be $x^\circ .$
Then, another angle is $\frac{\text{x}^\circ}{3}.$
Thus, $\text{x}+\frac{\text{x}}{3}=180^\circ$
$4\text{x}=180\times3\text{x}=\frac{180\times3}{4}=135^\circ$
Thus, the required angles are $135^\circ , 45^\circ .$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
Since, $PQ || CD$
$\therefore \angle \text{EFC}=\angle \text{FEQ}=37^\circ$ [Alternate angles]
Now, $\angle \text{AEQ}+\angle \text{FEQ}=\angle \text{AEF}$
$\Rightarrow \angle \text{AEQ}+37^\circ=95^\circ$
$\Rightarrow \angle \text{AEQ}=58^\circ$
Since, $PQ || AB$
$\therefore \angle \text{EAB}+\angle \text{AEQ}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \angle\text{EAB}+58^\circ=180^\circ$
$\Rightarrow \angle\text{EAB}=122^\circ$
$\angle\text{EAB}+\text{Reflex}\angle\text{EAB}=360^\circ$ [Complete angle]
$\therefore 122^\circ+(2\text{x})^\circ=360^\circ$
$\Rightarrow 2\text{x}=238$
$\Rightarrow \text{x}=119$
Hence, the correct answer is option $(d).$

cLet the required angle be $x$
Then,
$x = 3(180^\circ - x)$
$\Rightarrow x = 540^\circ - 3x$
$\Rightarrow 4x = 540^\circ $
$\Rightarrow x = 135^\circ $
Hence, the correct answer is option $(c).$

 From the given figure, we can say that.
$\angle\text{BOC}=\angle\text{EOF} 40^\circ=\angle\text{EOF}$ [vertycally opposite angles]
$\Rightarrow40^\circ=\angle\text{EOF}$
Since, sum of all the angles on a straight line is $180^\circ $
$\therefore\angle\text{BOC} +\angle\text{FOE}+\angle\text{EOD}=180^\circ$
$\Rightarrow 90^\circ+40^\circ+5\text{a}=180^\circ$
$\Rightarrow130^\circ+5\text{a}=180^\circ\Rightarrow5\text{a}=180^\circ-130^\circ$
$\Rightarrow5\text{a}=50^\circ$
$\Rightarrow\text{a}=\frac{50^\circ}{5}=10^\circ$

 We know that, the sum of all angles around a point is $360^\circ .$
$\therefore100^\circ+46^\circ+64^\circ+\text{x}=360^\circ$
$\Rightarrow210^\circ+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-210^\circ$
$\Rightarrow\text{x}=150^\circ$

 Construction: Draw a line $OE$ from the point $O$ parallel to $AB$ and $CD$

Since, $AB || OE$
$\therefore\angle \text{BAO}+\angle \text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 150^\circ+\angle \text{AOE}=180^\circ$
$\Rightarrow \angle \text{AOE}=30^\circ$
Again, $CD || OE$
$\therefore \angle \text{DCO}+\angle \text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 120^\circ+\angle \text{COE}=180^\circ$
$\Rightarrow \angle \text{COE}=60^\circ$
Now, $\angle \text{AOC}=\angle \text{AOE}+\angle \text{COE}$
$=30^\circ+60^\circ$
$=90^\circ$
Hence, the correct answer is option $(b).$

 Let the angle be $A$ and $B$
$A + B = 90^\circ ($complementary angles$)$
If $A = 17x, B = 13x ($Given$)$
$17x + 13x =90^\circ $
$30x = 90^\circ $
$x = 3^\circ $
$\therefore A = 17 \times 3^\circ = 51^\circ , B = 13 \times 3^\circ = 39^\circ $

 Complement of $30^\circ = 60^\circ $
Supplement of $60^\circ = 120^\circ $

Both Angles forming linear pair cannot be acute as they add up to form $180$ degrees.
Hence one angle can be acute and other be obtuse or both the angles can be right angles if they form linear pair.
Hence the above statement is false.

 Angle $= 230$ Complementary
$\angle=90-23^\circ=67^\circ$
Supplementary
$\angle=180-67^\circ=113^\circ$

 Complementary angle of $60^\circ = 90^\circ - 60^\circ = 30^\circ $

 Supplementary angle are two angles sum of $180^\circ $
Complementary angle are two angles sum of $90^\circ $
Supplementary angle $= 180 - 100 = 80^\circ $
Complementary angle $= 90 - 80 = 10^\circ $

Two angles whose sum is $180^\circ $ are said to be supplementary.
Given two angles are the ratio of $7 : 11.$
Let the two angles be $7x$ and $11x.$
So, $7x + 11x = 180^\circ $
$\Rightarrow 18x = 180^\circ $
$\Rightarrow\text{x}=\frac{180}{18}$
$x = 10$
Therefore, the two angles are $7x = 70$ and $11x = 110$

Since, two angles are supplementary their sum is $180^\circ $
$\angle1+\angle2=180^\circ$
$49^\circ+\angle2=180^\circ$ (As one of the angle is $49^\circ $
$\angle2=180^\circ-49^\circ$
$=131^\circ$

 Let $x$ and $y$ be supplementary angles
$\Rightarrow x + y = 180^\circ $
Let x be an angle which is $5$ times its supplement
$\Rightarrow x = 5y$
But y $= 180^\circ − x .......$ From $(i)$
$\Rightarrow x = 5 (180^\circ - x)$
$\Rightarrow x = 5 \times 180^\circ - 5x$
$\Rightarrow 6x = 5 \times 180^\circ $
$\Rightarrow x = 5 \times 30^\circ = 150^\circ $
Hence, $x = 150^\circ $

Two angles whose sum is $180^\circ $ are said to be supplementary.
Two angles whose sum is $90^\circ $ are said to be complementary.
Let the angle be $x.$
Given that $5 (90 - x) = 2(180 - x) - 24$
$\Rightarrow 450 - 5x = 360 - 2x - 24$
$\Rightarrow 5x - 2x = 450 - 360 + 24$
$\Rightarrow 3x = 474 - 360$
$\Rightarrow 3x = 114$
$\Rightarrow\text{x}=\frac{114}{3}\Rightarrow\text{x}=38$
Therefore, the angle is $38^\circ .$

 $\angle{\text{B}}=180^\circ-\angle{\text{A}}=180^\circ=118^\circ=62^\circ$
$\angle{\text{C}}=90^\circ-62^\circ=28^\circ.$

 Given two supplementary angles are in the ratio $3 : 2.$
Let the measurement of the angles be $3x$ and $2x.$
Two angles are said to be supplementary if they sum upto $180^\circ .$
Then we have, $3x + 2x = 180^\circ $
$5x = 180^\circ $ or, $x = 36^\circ .$
So the smaller angle is $36^\circ \times 2 = 72^\circ .$

In a pair of complimentary angles, sum of angles is $90^\circ $
$60^\circ + 24^\circ = 90^\circ $

 Given the angles of a triangle are in the ratio $2 : 3 : 7.$
Let the angles of triangle be $2x, 3x$ and $7x.$
Then according to the problem we get,
$2x + 3x + 7x = 180^\circ $
or, $12x = 180^\circ $
or, $x = 15^\circ .$
Then the largest angle is $ 7 \times 15^\circ = 105^\circ .$

 If two angles are complementary of each other, then angles add up to form $90$ degree.
The angles are less than $90.$
Hence, angles which are complementary of each other are acute angles.

 Complement of an angle $A = 90^\circ - A$
So, complement of angle
$3x - 8^\circ = 90^\circ - (3x - 8^\circ )$
$= 98^\circ - 3x$

 Only one acute and one obtuse can be formed on a same side of straight line.
Say, if a angle is $60^\circ$ then another angle will be $(180^\circ - 60^\circ ) = 120^\circ ,$
So one acute and one obtuse angle can be formed.

 Let, $\alpha$ be the larger angle and \beta be the smaller angle.
if, two angles are complementary then their sum is equal to $90^\circ $
So, $\alpha+\beta=90^\circ....(1)$
According to question, $\alpha=2\beta.....(2)$
So, $2\beta+\beta=90^\circ ($from eqn$(1)$ and eqn $(2))$ or, $3\beta=90^\circ$ or $\beta=30^\circ.$
Therefore, the smaller angle $= 30^\circ $

When a transversal intersects two parallel lines, the angle made on the interior same side is $180$ degrees.
So, both are facts but reason does not explain assertion correctly.

 Two angles are said to be supplementary, if their sum is $180^\circ .$ Also, if two angles are vertically opposite, then they are equal.
Therefore, angles given in option $(b)$ are supplementary as well as vertically opposite.

 Let the angles be $2x$ and $3x$
Now sum of interior angles on same side of transversal intersecting two parallel lines is $180^\circ $
$\Rightarrow 2x + 3x = 180^\circ $
$\Rightarrow 5x = 180^\circ $
$\Rightarrow x = 36^\circ $
So the angles are $2x = 2 \times 36^\circ = 72^\circ $
$3x = 3 \times 36^\circ = 108^\circ $
So the smaller angle is $72^\circ .$

Two angles are Comple mentary when they add upto form $90$ degrees (Right Angle).
If one angle $= 77$
Let the other angle be $x$
Hence $x = 90 − 77$
$= 13$
Hence complementarty angle of the following angle is $13$

Let each vertical angle be $x$
Now the sum of vertical angles is $360^\circ $
$\Rightarrow\text{x}+\text{x}+\text{x}+\text{x}=360^\circ$
$\Rightarrow4\text{x}=360^\circ$
$\Rightarrow\text{x}=\frac{360^\circ}{4}=90^\circ$

Let the supplement be $x$ Sum of supplementary angles is $180^\circ $
$\therefore x + 36^\circ = 180^\circ $
$\Rightarrow x = 144^\circ $
Difference between supplement and given angle
$= 144 − 36 = 108$

Given angle is $72.5$ or $72\frac{1^\circ}{2}$
Let the other angle $= x$ Sum of complementary angles
$=90^\circ\Rightarrow\text{x}+72\frac{1^\circ}{2}=90^\circ$
$\Rightarrow\text{x}=17\frac{1^\circ}{2}$

From the above figure, it is clear that the angle between North and West is $90^\circ $ and South and East is $90^\circ .$
$\therefore$ Sum of these two angles $= 90^\circ + 90^\circ = 180^\circ $
Hence, the two angles are supplementary, as their sum is $180^\circ .$

 Two angles the sum of whose measure is $90^\circ $ is called Complimentary angles.

 Let the angles be $x$ and $9x$ Sum of complementary angles is $90^\circ $
$\Rightarrow x + 9x = 90^\circ $
$\Rightarrow 10x = 90^\circ $
$\Rightarrow x = 9^\circ $
So the angles are $1 \times 9^\circ = 9^\circ $
$9 \times 9^\circ = 81^\circ $

 Let the supplement be $x$
If angles are supplementary then their sum is $180^\circ $
$\Rightarrow x + 100^\circ = 180^\circ $
$x =180^\circ - 100^\circ $
$x = 80^\circ $

 
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
$\angle \text{FCD}+\text{Reflex}\angle \text{FCD}=360^\circ$ (Complere angle)
$\Rightarrow \angle \text{FCD}+273^\circ=360^\circ$
$\Rightarrow \angle \text{FCD}=87^\circ$
Since, $PQ || CD$
$\therefore \angle \text{QFC}+\angle \text{FCD}=180^\circ$ (Angles on the same side of a transversal line are supplementary)
$\Rightarrow \angle \text{QFC}+87^\circ=180^\circ$
$\Rightarrow \angle \text{QFC}=93^\circ$
Now, $\angle \text{ABF}=\angle \text{BFQ}$ (Corresponding angles)
$=\angle \text{BFC}+\angle \text{QFC}$
$=54^\circ+93^\circ$
$=147^\circ$
$\therefore \text{x}^\circ=147^\circ$
$\Rightarrow \text{x}=147$
Hence, the correct answer is option $(c).$

 Suppose a circle is drawn passing through all the vertices of the triangle $ABC$ with centre at $O$ (orthocenter). thus the angle formed at the orhtocenter is the supplement of the angle at the vertex.
$\angle\text{BOC}+\angle\text{BAC}=180^\circ$
So, $\angle\text{BOC}$ and $\angle\text{BAC}$ are supplementary.

 Let the angles be $4x$ and $5x$ Angles are supplementary
$\therefore 4x + 5x = 180^\circ $
$\Rightarrow 9x = 180^\circ $
$\Rightarrow x = 20^\circ $
So the angles are
$4 \times 20^\circ = 80^\circ $
$5 \times 20^\circ = 100^\circ $

 Since $\angle\text{POR}$ and $\angle\text{QOS}$ are in the ratio $1 : 5$ Let angles will be $x$ and $5x,$ respectively. We know that, the sum of angles forming linear pair is $180^\circ $
$\therefore \angle\text{POR}+ \angle\text{ROS}+\angle\text{QOS}=180^\circ$
$\Rightarrow\text{x}+90^\circ+5\text{x}=180^\circ$
$\Rightarrow6\text{x}=180^\circ-90^\circ$
$\Rightarrow6\text{x}=90^\circ\Rightarrow \text{x}=\frac{90^\circ}{6}$
$\text{x}=15^\circ$
$\therefore\angle\text{QOS}=5\text{x}=5\times15^\circ$
$\angle\text{QOS}=75^\circ$

  Since, $POQ$ is a line.
Here, $\angle\text{POR}$, and $\angle\text{QOR}$ from a liner pair.
$\therefore\angle\text{POR}+\angle\text{QOR}=180^\circ [ \therefore$ Sum of the liner pair is $180^\circ ]$
$\Rightarrow100^\circ+\text{a}=180^\circ$
$\Rightarrow\text{a}=180^\circ-100^\circ=80^\circ$

 It is given that, $POQ$ is a line. Since, sum of all the angles on a straight line is $180^\circ .$
Therefore, $\text{x}+2\text{y}+3\text{y}=180^\circ$
$\Rightarrow\text{x}+5\text{y}=180^\circ$ $[\because\text{x}=30^\circ,\text{given}]$
$\Rightarrow 30^\circ+5\text{y}=180^\circ$
$\Rightarrow 5\text{y}=180^\circ-30^\circ$
$\Rightarrow 5\text{y}=150^\circ$
$\Rightarrow\text{y} = \frac{150^{\circ}}{5}$$$
$\Rightarrow \text{y}=30^\circ$
$\therefore\angle\text{QOR}=3\text{y}=3\times30^\circ=90^\circ$

$\angle \text{BPE}=\angle \text{APQ}=(5\text{x}-10)^\circ$ [Vertically opposite angles]
Since, $AB || CD$
$\therefore \angle \text{APQ}+ \angle \text{CQP}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow(5\text{x}-10)^\circ+(3\text{x}-10)^\circ=180^\circ$
$\Rightarrow 8\text{x}-20=180$
$\Rightarrow 8\text{x}=200$
$\Rightarrow \text{x}=25$
$\therefore \angle \text{BPE}=(5\times 25-10)^\circ=115^\circ$
Now, $\angle \text{BPE}=\angle \text{DQP}=115^\circ$ [Corresponding angles]
Hence, the correct answer is option $(c).$

We know acute angle $ < 90^\circ $
Let us take an example.
$\angle\text{x}=45^\circ.....\text{x}$ is an acute angle.
Supplement of $x$ is $180^\circ − 45^\circ = 135^\circ $
It is an Obtuse angle Obtuse angle $ > 90^\circ $.

 Five sixth of right angle $=\frac{5}{6}\times90^\circ=75^\circ$
Let the supplement be $x$
$\therefore x + 75^\circ = 180^\circ $
$\Rightarrow x = 180^\circ − 75^\circ $
$\Rightarrow x = 105^\circ $

 Let the angle be $x^\circ .$ Then, the complement of $x$ will be $(90 - x)^\circ .$
Given, complement of $x^\circ $ is $79^\circ .$
$\therefore(90-\text{x})^\circ=79^\circ$
$\Rightarrow\text{x}^\circ=90^\circ-79^\circ=11^\circ$
Therefore, the required angle is $11^\circ .$
Note Sum of the complementary angles is $90^\circ .$

$PQ$ and $RS$ intersect at $O.$ then,
$\angle\text{POS}=\angle\text{QOR}$ (opposite angles)
$\angle\text{SOQ}=\angle\text{POR}$ (opposite angles)
Given, $\angle\text{POS}=2\angle\text{SOQ}$
Sum of all angles $= 360$
$\angle\text{POS}+\angle\text{SOQ}+\angle\text{QOR}+\angle\text{ROP}=360$
$6\angle\text{SOQ}=360$
$\angle\text{SOQ}=60$
Hence, the four angles $= 60^\circ , 60^\circ , 120^\circ , 120^\circ .$

Answer is option $A$
If thetransversalcrosses two parallellines.
Each pair of interior angles are inside the parallel lines, and on the same side of the transversal.
are supplementary $($add to $180$ degrees$)$

 Let one of the angle be $x.$ Since, the difference between the two angles is $30^\circ $,
then the other angle will be $(x – 30^\circ ).$
Also, the two angles are complementary, so their sum is equal to $90^\circ .$
$\therefore\text{x}+\text{(x}-30^\circ)=90^\circ$
$\Rightarrow\text{x}+\text{x}-30^\circ=90^\circ$
$\Rightarrow2\text{x}=90^\circ+30^\circ$
$\Rightarrow2\text{x}=120^\circ$
$\Rightarrow\text{x}=\frac{120^\circ}{2}$
$\Rightarrow\text{x}=60^\circ$
$\therefore$ Required angles are $60^\circ $ and $(60^\circ - 30^\circ ),$ i.e. $60^\circ $ and $30^\circ $

 Let the angles be $2x$ and $3x$ Sum of angles on the same side of transversal intersecting two parallel lines is $180^\circ $
$\Rightarrow 2x + 3x = 180^\circ $
$\Rightarrow 5x = 180^\circ $
$\Rightarrow x = 36^\circ $
So the angles are $2x = 2 \times 36^\circ = 72^\circ $
$3x = 3 \times 36^\circ = 108^\circ $
So the smaller angle is $72^\circ $

 Since, sum of all the angles on a straight line is $180^\circ .$
Therefore, $6\text{y}+\text{y}+2\text{y}=180^\circ$
$\Rightarrow9\text{y}=180^\circ$
$\Rightarrow\frac{180^\circ}{9}$
$\therefore\text{y}=20^\circ$

Sum of angles of triangles $= 180^\circ $
$2x + 3x - 5 + 4x - 13 = 180^\circ $
$9x - 18 = 180^\circ $
$9x = 198^\circ $
$x = 22^\circ $

$\angle \text{AOC}+\angle\text{BOC}=180^\circ$ [$\because$ Linear pair angles]
$\Rightarrow \text{y}^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{y}+\text{x}=180$
$\Rightarrow \frac{4\text{x}}{5}+\text{x}=180$ $\big[\because 4\text{x}=5\text{y}\Rightarrow \text{y}=\frac{4\text{x}}{5}\big]$
$\Rightarrow 4\text{x}+5\text{x}=180\times 5$
$\Rightarrow 9\text{x}=180\times 5$
$\Rightarrow \text{x}=100$
Hence, the correct answer is option $(a).$

It is given that the angles are in the ratio of $1 : 2.$
Let the angles will be $x$ and $2x.$
Also, the two angles are supplementary, i.e. their sum is equal to $180^\circ .$
$\therefore \text{x}+2\text{x}=180^\circ$
$\Rightarrow3\text{x}=180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{3}$
$\Rightarrow\text{x}=60^\circ$
Hence, the required angles are $60^\circ $ and $2 \times 60^\circ ,$ i.e. $60^\circ $ and $120^\circ $
$\therefore$ Bigger of the two angles is $120^\circ .$

 Supplementary of $\angle \text{A}=180^\circ-\angle \text{A}$
Now,
$\angle \text{A}+30^\circ=2(180^\circ-\angle \text{A})$
$\Rightarrow \angle \text{A}+30^\circ=360^\circ-2\angle \text{A}$
$\Rightarrow 3\angle \text{A}=360^\circ-30^\circ$
$\Rightarrow 3\angle \text{A}=330^\circ$
$\Rightarrow \angle \text{A}=110^\circ$
Hence, the correct answer is option $(b).$

Sum of angles of triangles $= 180^\circ $
$2x + 3x - 5 + 4x - 13 = 180^\circ $
$9x - 18 = 180^\circ $
$9x = 198^\circ $
$x = 22^\circ $

 
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
$\angle \text{QFC}+\angle \text{ECD}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \angle \text{QEC}+56^\circ=180^\circ$
$\Rightarrow \angle \text{QEC}=124^\circ$
Now, $\angle \text{BEQ}+\angle \text{QEC}=\angle \text{BEC}$
$\Rightarrow \angle \text{BEQ}+124^\circ=158^\circ$
$\Rightarrow\angle \text{BEQ}=34^\circ$
Now, $\angle \text{ABE}=\angle \text{BEQ}=34^\circ$ [Corresponding angles]
$\therefore \text{x}^\circ=34^\circ$
$\Rightarrow \text{x}=34$
Hence, the correct answer is option $(a).$

It is given that, angles $P$ and $O$ are supplementary. Hence, the sum of $P$ and $O$ will be $180^\circ $
$\therefore\angle\text{P}=\angle\text{Q}=180^\circ$
$\Rightarrow60^\circ=\angle\text{Q}=180^\circ$ $[\because\angle=60^\circ,\text{given}]$
$\angle\text{Q}=180^\circ-60^\circ $
$\angle\text{Q}=120^\circ$

From the given Figure it is clear that, $\angle\text{a}=\angle\text{b}$ and $\angle\text{c}=\angle\text{d}$.
[vertically opposite angles]
Also, $\angle\text{a}=\angle\text{b}=180^\circ$
And $\angle\text{c}=\angle\text{d}=180^\circ$[Liner pair]

Draw a line $l$ parallel to $QP.$


$\therefore \angle \text{PQT}=\text{x}$
$\Rightarrow \text{x}=60^\circ$ [Alternate interior angles]
Also, $\angle\text{RST}=\text{y}$
$\Rightarrow \text{y}=30^\circ$ [Alternate interior angles]
Now, $\text{a}=\text{x}+\text{y}$
$\Rightarrow\text{a}=60^\circ+30^\circ$
$\Rightarrow\text{a}=90^\circ$

Given angle is $= 108.5^\circ $
Let the angle supplementary with above angle be $x.$
Now, sum of two supplementary angles $= 180^\circ $
$\Rightarrow x + 108.5^\circ = 180^\circ $
$\Rightarrow x = 71.5^\circ .$

Since, $PQ || RS$ and $QR$ is transversal.
$\therefore\angle\text{PQR}=\angle\text{SRQ}$ [Alternate interior angles]
$\Rightarrow\angle\text{SRQ}=85^\circ$
Also, $ST || QR$ and $RS$ is transversal.
$\therefore\angle\text{SRQ}=\angle\text{RST}$ [Alternate interior angles]
$\Rightarrow\angle\text{RST}=85^\circ$
Now, $\angle\text{RST}+\text{a}=180^\circ$ [Liner pair]
$\Rightarrow \text{a}= 180^\circ-\angle\text{RST}$
$\Rightarrow \text{a}=180^\circ-85^\circ$
$\Rightarrow \text{a}=95^\circ$ $[\because\angle\text{RST}=85^\circ]$

Two angles are supplementary if their sum is $180^\circ $
If one angle is $\frac{2}{5}$ of a right angle, then other angle is
$180-\Big(\frac{2}{5}\times90\Big)$
$=180-(2\times18)$
$=180-36=144^\circ$

Two angles are complementary if their sum is $90^\circ .$
If one angle is $\frac{1}{4},$ then other angle is $\frac{3}{4}\times90=67.5$

Let the required angle be $x,$ then its supplement $= (180 - x)$
Given that $x = (180 - x) + 20$
$⇒ 2x = 200$
$\Rightarrow x = 100^\circ $

Let the required angle be $x$
Therefore, $x = 2 (180 - x)$
$\Rightarrow x = 360 - 2x$
$\Rightarrow 3x = 360$
$\Rightarrow x = 120^\circ $

Let the required angle be $x.$
Then, its supplement will be $(180^\circ – x).$
It is given that, the angle is four times its supplement.
Therefore, $\text{x}=4(180^\circ-\text{x})$
$\Rightarrow\text{x}=4\times180^\circ-4\text{x}$
$\Rightarrow \text{x}+4\text{x}=720^\circ$
$\Rightarrow 5\text{x}=720^\circ$
$​​​​\Rightarrow\text{x}=\frac{720^\circ}{5}$
$\Rightarrow \text{x}=144^\circ$
Hence, the required angle is $144^\circ .$

Let another angle be $x + 25^\circ = 180^\circ $
$x = 180^\circ - 125^\circ $
$x = 55^\circ $
Hence, another angle $= 55^\circ $

 Given, $2\angle\text{A}=3\angle\text{B}=6\angle\text{C},$
$2\angle\text{A}=6\angle\text{C}\angle\text{A}=3\angle\text{C}$
$3\angle\text{B}=6\angle\text{CB}=2\angle\text{C}$
Now, $\angle\text{A}=\angle\text{B}=\angle\text{C}=180^\circ$
$3\angle\text{C}+2\angle\text{C}+\angle\text{C}=180^\circ$
$6\angle\text{C}=180^\circ$
$\angle\text{C}=30^\circ$
Small angle $= 30^\circ $

 Answer is option $A$
If a ray stands on a line, then the sum of two adjacent angles so formed is $180.$
Conversely if the sum of two adjacent angles is $180,$ then a ray stands on a line (i.e., the non-common arms form a line).

 Since, $PQ || ST,$ then $PO$ will also parallel to $ST.$
Now, $PO || ST$ and $OS$ is transversal.
Therefore,
$x = 85^\circ [$Alternate interior angles$]$
Now, $y + 130^\circ = 180^\circ [$Liner pair$]$
$\Rightarrow y = 180^\circ - 130^\circ $
$\Rightarrow y = 50^\circ $
$\therefore x + y = 85^\circ + 85^\circ = 135^\circ $ 

Let the angles be $x$ and $y$ Given $x − y = 44^\circ .....(i)$
Sum of supplementary angles is $x + y = 180^\circ ......(ii)$
Solving $(i)$ and $(ii)$
$\Rightarrow x = 112^\circ , y = 68^\circ $

Let the required angle be $x^\circ .$ It is given that $x^\circ$ makes a linear pair with $61^\circ $
$\therefore  x + 61^\circ = 180^\circ [ \therefore$ sum of angles forming linear pair is $180^\circ ]$
$\Rightarrow x = 180^\circ - 61^\circ = 199^\circ $

 Let the required angle be $x$
Now, complementnary of the required angle $= 90^\circ - x$
Then,
$\text{x}+\frac{1}{2}(90^\circ-\text{x})=75^\circ$
$\Rightarrow2\text{x}+90^\circ-\text{x}=150^\circ$
$\Rightarrow \text{x}=150-90^\circ$
$\Rightarrow \text{x}=60^\circ$
Hence, the correct answer is option $(c).$

 Let the supplementary angle be $x$ Sum of supplementary angles is $180^\circ $
$\Rightarrow x + 120^\circ = 180^\circ $
$\Rightarrow x = 180^\circ - 120^\circ $
$\Rightarrow x = 60^\circ $

Two adjacent angles whose sum is $180^\circ $ is called linear pair.

Supplement of $150^\circ $ is $30^\circ $ and complement angle of $30^\circ $ is $60^\circ .$

 Let the angles be $xx$ and its supplement be $y = 180 - x$
Given: $\text{x}=\frac{1}{5}\text{y}$
$\Rightarrow\text{x}=\frac{1}{5}\times(180-\text{x})$
$\Rightarrow5\text{x}=180-\text{x}$
$\Rightarrow6\text{x}=180$
$\therefore\text{x}=\frac{180}{6}$
$\therefore\text{x}=30^\circ$

$\perp$ represents Perpendicularity.
Here Line $PQ$ is perpendicular to Line $RS$
$\therefore\overline{\text{PQ}}\perp\overline{\text{RS}}$

Supplementary angles add up to form $180$
Let one angle be $x$ and other be $180 - x$
Hence, $x - (180 − x) = 48 ....($Given$)$
$⇒ x - 180 + x = 48$
$⇒ 2x = 48 + 180 = 228$
$\Rightarrow\text{x}=\frac{288}{2}=114$
Hence, other angle $= 180 - x = 180 - 114 = 66$
Two angles are $114$ and $66.$

Since, $AB || EG$
$\therefore \angle \text{ABG}+\angle \text{EGB}=180^\circ$ (Angles on the same side of a transversal line are supplementary)
$\Rightarrow 110^\circ+\angle \text{EGB}=180^\circ$
$\Rightarrow \angle \text{EGB}=70^\circ$
Again, $CD || GF$
$\therefore \angle \text{DCG}+\angle \text{FGC}=180^\circ$ (Angles on the same side of a transversal line are supplementary)
$\Rightarrow 100^\circ+\angle \text{FGC}=180^\circ$
$\Rightarrow \angle \text{FGC}=80^\circ$
Now, $\angle \text{EGB}+\angle \text{BGC}+\angle \text{FGC}=180^\circ$
$\Rightarrow 70^\circ+\text{x}^\circ+80^\circ=180^\circ$
$\Rightarrow 150^\circ+ \text{x}^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=30^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(c).$

 Two angles are said to be complimentary angles if their sum is $90^\circ .$
Let $x$ denote the required angle. Then its complimentary angle is $90 − x.$
It is given that,
$x = 8 \times (90 − x)$
$\Rightarrow x = 720 − 8x$
$\Rightarrow 9x = 720$
$\Rightarrow x = 80^\circ $

 The two adjacent angles add up to $140 .$
Therefore the sum of the two should give $140.$
Let the larger be $x$
then the smaller is $x - 28$
$x + x - 28 = 140$
$2x - 28 = 140$
$2x = 140 + 28 = 168$
$2x = 168 ($divide both sides by two$)$
$x = 84 ($the larger angle$)$
$84 - 28 = 56 ($the smaller angle$)$
The two angles are $84$ and $56.$

 Let the required angle be $x,$ then its complement $= (90 - x)$
Given that $x = (90 - x) + 80$
$\Rightarrow \text{x}=\frac{170}{2}$
$\Rightarrow \text{x}=85^\circ$

 From the given figure it is clear that,
$2a + 5b = 180^\circ [$Liner pair$]$
$\Rightarrow 2 \times 40^\circ + 5b$ $180^\circ [\because a = 40^\circ ]$
$\Rightarrow 80^\circ + 5b = 180^\circ $
$\Rightarrow 5b = 180^\circ - 80^\circ $
$\Rightarrow 5b = 100^\circ $
$\Rightarrow\text{b}=\frac{100^\circ}{5}$
$\Rightarrow b = 20^\circ $