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

 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 $x = (90 - x) + 80$
$\Rightarrow \text{x}=\frac{170}{2}$
$\Rightarrow \text{x}=85^\circ$