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

 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 $