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 $(a).$