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CBSE Questions for Class 9 Maths Lines And Angles Quiz 5 - MCQExams.com
CBSE
Class 9 Maths
Lines And Angles
Quiz 5
If angle $$P$$ and angle $$Q$$ are supplementary and the measure of angle $$P$$ is $$60^o$$, then the measure of angle $$Q$$ is:
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$$120^o$$
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$$60^o$$
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$$30^o$$
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$$20^o$$
Explanation
Option (a) is correct.
Given that $$\angle P +\angle Q=180^o$$
$$\Rightarrow 60^o +\angle Q=180^o$$
$$\Rightarrow \angle Q=180^o -60^o$$
$$\Rightarrow \angle Q=120^o$$
Two acute angles are congruent.
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True
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False
Explanation
False
Two acute angles may vary from $$0^{\mathring{}}$$ to $$89^{\mathring{}}$$. Hence, acute angles having different measures are not congruent.
Sum of any two angles of a triangle is always greater than the third angle.
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True
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False
Explanation
Sum of any two angles of a triangle is may or may not be greater than the third angle.
If the sum of two angles is equal to an obtuse angle, then which of the following is not possible?
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One obtuse angle and one acute angle
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One right angle and one acute angle
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Two acute angles
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Two right angles
Explanation
Option [D] is the correct answer.
The sum of two right angles is exactly equal to $${180}^{o}$$, which is a straight angle not an obtuse angle.
In the figure, if OP||RS, $$\angle OPQ=110^0 and \angle QRS =130^0$$, then $$\angle PQR$$ is equal to
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$$40^0$$
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$$50^0$$
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$$60^0$$
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$$70^0$$
Explanation
As
OP||RS,
$$\angle$$ PQR $$ =180 -(180-110)-(180-130) $$=$$60$$
If m$$\angle$$ $$XYZ = 86^o$$ and m$$\angle$$ $$XZY = 23^o$$. What is m$$\angle$$ $$YXZ$$ in the triangle?
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$$70^o$$
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$$71^o$$
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$$72^o$$
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$$73^o$$
Explanation
By angle sum property, the sum of angles is $$180^o$$.
$$86^o + 23^o +$$ m$$\angle$$ $$YXZ = 180^o$$
m$$\angle$$ $$YXZ = 180^o - 109^o$$
m$$\angle$$ $$YXZ = 71 ^o$$.
Find $$x + y + z$$
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$$ x+y+z = 180^{0}$$
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$$ x+y+z = 360^{0}$$
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$$ x+y+z = 480^{0}$$
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$$ x+y+z = 540^{0}$$
Explanation
Here, by exterior angle property, $$\angle y$$ being an exterior angle, will be the sum of opposite interior angles.
Then, $$\angle y= 90^o + 30^o = 120^{\circ}$$.
We know, by straight angle property, sum of angles on a straight line $$= 180^o$$.
Then, $$\angle z +30^o = 180^o$$
$$\implies$$ $$\angle z = 150^{\circ}$$.
Also,
$$\angle x + 90^o = 180^o$$
$$\implies$$
$$\angle x = 90^{\circ}$$.
Therefore, $$ x +y +z = 120^o +150^o +90^o = $$ $$360^{\circ}$$.
Therefore, option $$B$$ is correct.
In the given figure, $$\angle QPB$$ is,
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$$60^{\circ}$$
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$$45^{\circ}$$
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$$30^{\circ}$$
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$$15^{\circ}$$
Explanation
In the given figure,
$$\angle APR + \angle RPQ + \angle QPB = 180^{\circ}$$
$$2x + 3x+x = 180^{\circ}$$
$$6x = 180^{\circ}$$
$$x = 30^{\circ}$$
If two supplementary angles are in the ratio 2 : 7, then the angles are :
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$$35^{\circ}, 145^{\circ}$$
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$$70^{\circ}, 110^{\circ}$$
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$$40^{\circ}, 140^{\circ}$$
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$$50^{\circ}, 130^{\circ}$$
Explanation
$$\textbf{ Hint: Sum of supplementary angles is}$$ $$180^\circ$$
$$\textbf{Step1: Evaluate using addition of ratio}$$
$$\text{Given}$$
$$\text{ratio}$$ $$=2:7$$
$$\text{ Let angles as }$$ $$2x$$ $$\text{and}$$ $$7x$$
$$\text{As we know that sum of supplementary angle is }180$$
$$\Rightarrow 2x+7x=180$$
$$\Rightarrow 9x=180$$
$$\Rightarrow x=20$$
$$2x\Rightarrow 2(20)=40$$
$$7x\Rightarrow 7(20)=140$$
$$\text{ Hence, the angles are}$$ $$40^\circ$$ $$140^\circ.$$
$$\textbf{Option C is correct.}$$
In the figure, if $$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = k\times $$ right angle, then $$k$$ is :
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$$2$$
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$$3$$
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$$4$$
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$$5$$
Explanation
$$\Rightarrow$$ In $$\triangle ABC$$
$$\Rightarrow$$ $$\angle A+\angle B+\angle C=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] --- ( 1 )
$$\Rightarrow$$ In $$\triangle DEF$$
$$\Rightarrow$$ $$\angle D+\angle E+\angle F=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] ---( 2 )
$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^\circ+180^\circ$$ [Adding ( 1 ) and ( 2 )]
$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^\circ$$
$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times (right\, angle)$$.
$$\therefore$$ $$k=\dfrac{360^o}{90^o}$$
$$\therefore$$ $$k=4$$
Two angles are called adjacent if
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they have a common vertex
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they have a ray in common
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their other arms lie on the opposite sides of the common arm
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all the above
Explanation
We know that, adjacent angles are angles that have a common vertex and a common side, but do not overlap. Also, their other arms lie on the opposite side of the common arm.
So, all the given options are correct.
Hence, option D is the answer.
If one of the angles of a triangle is $$130^0$$, then the angle between the bisectors of the other two angles can be
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$$50^0$$
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$$65^0$$
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$$145^0$$
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$$155^0$$
Explanation
Let $$x$$ and $$y$$ be the bisected angles.
So, in the original triangle,
sum of angles $$=180^o$$
Therefore, $$2x+2y+130=180$$
$$\Rightarrow 2(x+y)=50$$
$$\Rightarrow x+y=25$$
In the smallest triangle, consisting of original side opposite $$130^o$$.
Therefore, $$25^o+$$ Angle between bisectors $$=180^o$$
$$\Rightarrow $$ Angle between bisectors of other two angles $$=155^o$$.
If the sum of two adjacent angles is $$100^{\circ}$$ and one of them is $$35^{\circ}$$, then the other is :
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$$70^{\circ}$$
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$$65^{\circ}$$
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$$135^{\circ}$$
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$$145^{\circ}$$
Explanation
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 $$65^{\circ}$$
The angle which exceeds its complement by $$20^{\circ}$$ is:
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$$45^{\circ}$$
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$$55^{\circ}$$
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$$70^{\circ}$$
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$$110^{\circ}$$
Explanation
Let the required angle be $$x$$.
We knows, complementary angles $$=$$ Sum of two angles is $$90^o$$.
$$\therefore x = (90^o - x ) + 20^o $$
$$\therefore$$ $$x = 90^o - x + 20^o$$
$$\therefore$$
$$ 2x = 110^o$$
$$\therefore x = 55^o$$.
Hence, option $$B$$ is correct.
From the adjoining figure, calculate the values of $$a$$.
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$$46^o$$
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$$38^o$$
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$$56^o$$
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none of the above
Explanation
We know, by angle sum property, the sum of angles of triangle $$= 180^o$$
Then, $$a + 110^o + 32^o = 180^o $$
$$\implies$$ $$a + 142^o = 180^o $$
$$a =180^o-142^o=38^{\circ}$$.
Thus measure of angle $$a$$ is $$38^o$$.
Hence, option $$B$$ is correct.
In the given figure, AB // CD, EH // BC, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of
$$\angle AEF$$ .
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$$36^o$$
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$$40^o$$
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$$46^o$$
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none of the above
Explanation
Given, $$\angle DGH = 40^{\circ}$$.
Since
$$AB \parallel CD$$,
$$\implies$$ $$AE \parallel DG$$.
By corresponding angles axiom,
$$\angle AEF = \angle DGH = 40^{\circ}$$.
Hence, option $$B$$ is correct.
In the given figure, $$ABC$$ is a triangle, side $$CB$$ is produced to $$E$$ and $$\angle A$$: $$\angle B$$: $$\angle C$$ $$= 2: 1: 3$$. Find $$\angle DBE$$, if $$DB$$ is perpendicular to $$AB$$.
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$$66^o$$
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$$46^o$$
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$$60^o$$
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none of the above
Explanation
In triangle $$ABC$$, $$\angle A : \angle B : \angle C = 2: 1: 3$$
Let the angles be $$2x, x, 3x$$
By angle sum property, the sum of the angles $$2x + x+ 3x = 180^o$$
$$\implies$$ $$6x = 180^o$$
$$\implies$$
$$x = 30^o$$.
Hence, $$\angle B = 30^o$$.
Now,
$$\angle DBE + \angle DBA + \angle ABC = 180^o$$ ...[Straight angle property]
$$\angle DBE + 90^o + 30^o = 180^o $$
$$\angle DBE = 180^o - 120^o $$
$$\angle DBE = 60^{\circ}$$.
Therefore, option $$C$$ is correct.
From the given figure, we can find that $$\angle ABC$$ $$=47^o$$.
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True
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False
Explanation
Given, $$\angle BAC = 53^o$$, $$\angle ACD = 100^o$$.
$$\angle ACD = \angle BAC + \angle ABC$$ ...(By exterior angle property, sum of interior opposite angles is equal to exterior angle)
$$\implies$$ $$100^o = 53^o + \angle ABC$$
$$\implies$$
$$\angle ABC = 100^o-53^o=47^{\circ}$$.
Hence, the statement is true.
Therefore, option $$A$$ is correct.
The angles of a triangle are in the ratio 2: 1:Is the triangle right-angled triangle?
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0%
True
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False
Explanation
The angles of the triangle are in the ratio, $$2:1: 3$$.
Let the angles be $$2x, x $$ and $$3x$$.
By angle sum property, sum of the angles $$= 180^o$$
$$2x + x+ 3x = 180^o$$
$$\implies$$ $$6x = 180^o$$
$$\implies$$
$$x = 30^o$$ .
Hence, the angles will be $$x=30^o, \ 2x=60^o$$ and $$3x=90^o$$.
Since, one of the angles is $$90^o$$, the triangle is a right angled triangle.
Hence, the statement is true and option $$A$$ is correct.
Two lines that are respectively parallel to two intersecting lines, intersect each other.
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True
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False
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Ambiguous
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Data Insufficient
Angles forming a linear pair can both be acute angles.
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True
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False
0%
Ambiguous
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Data Insufficient
Explanation
Answer is option B (False)
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.
Find the angle which is
$$80^o$$ more than its complement.
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$$75$$
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$$85$$
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$$95$$
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$$100$$
Explanation
Let the required angle be $$ x$$.
Then, its complement $$=(90^o-x)$$.
Given, the angle is $$80^o$$ more than its complement.
Then, $$ x=(90^o-x)+80^o$$
$$\Rightarrow x+x=90^o+80^o$$
$$ \Rightarrow 2x=170^o\\ \Rightarrow x=\dfrac { 170 }{ 2 } ^o\\ \Rightarrow x={ 85 }^{ o }$$.
Therefore, option $$B$$ is correct.
In the given figure, AB // CD, EH // BC, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of
$$\angle AFG$$.
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$$96^o$$
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$$86^o$$
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$$100^o$$
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none of the above
Explanation
Given that $$\angle BAC=60^o$$ and $$\angle DGH=40^o$$.
$$\implies$$
$$\angle EAF=60^o$$.
Also, given, $$AE \parallel DG$$.
Now, $$\angle AEF = \angle DGH = 40^{\circ}$$ (Corresponding angles).
Then, $$\angle AFG = \angle AEF + \angle EAF$$ (Exterior angle property)
$$\implies$$
$$\angle AFG = 40^{\circ} + 60^{\circ}$$
$$\implies$$
$$\angle AFG = 100^{\circ}$$.
Therefore, option $$C$$ is correct.
If two adjacent angles are equal, then each angle measures $$90^o$$.
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True
0%
False
0%
Ambiguous
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Data Insufficient
Explanation
The answer is $$B$$
Two adjacent angles measure $$90^o$$,
only when the lines are perpendicular to each other.
hence the above statement is False.
In the figure, given AE // BD and AC // ED: find $$\angle b$$.
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$$65^o$$
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$$36^o$$
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$$45^o$$
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none of the above
Explanation
In the given figure, $$\angle EDC = 65^o$$ (Vertically opposite angles).
Given, $$AC \parallel ED$$.
Thus, $$\angle b = \angle EDC = 65^o$$ (Corresponding angles on parallel lines are equal).
Therefore, option $$A$$ is correct.
Find the angle which is $$60^o$$ more than its complement.
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$$55$$
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$$75$$
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$$85$$
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$$60$$
Explanation
Let the required angle be $$ x$$.
Then, its complement $$=(90^o-x)$$.
Given, the angle is $$60^o$$ more than its complement.
Then, $$ x=(90^o-x)+60^o$$
$$\Rightarrow x+x=90^o+60^o$$
$$ \Rightarrow 2x=150^o\\ \Rightarrow x=\dfrac { 150 }{ 2 } ^o\\ \Rightarrow x={ 75 }^{ o }$$.
Therefore, option $$B$$ is correct.
In the given figure, $$AB || CD$$, $$EH || BC$$, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of
$$\angle CFG$$.
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$$60^o$$
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$$80^o$$
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$$56^o$$
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none of the above
Two supplementary angles differ by $$48^o$$. Then find these angles.
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$$36^o, 84^o$$
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$$46^o, 94^o$$
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$$56^o, 104^o$$
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$$66^o, 114^o$$
Explanation
We know that sum of supplementary angles is $$ 180^o$$.
Let one angle be $$x$$ and other be $$ 180^o - x$$
Hence, $$x - (180^o- x) =48^o$$ ....(Given)
$$ \Rightarrow x- 180^o+ x= 48^o$$
$$ \Rightarrow 2x= 48^o+ 180^o= 228^o$$
$$\Rightarrow x= \dfrac { 228^o }{ 2 } = 114^o$$
Hence, other angle $$= 180^o- x= 180^o- 114^o= 66^o$$
Two angles are $$ 114^o$$ and $$66^o$$.
The angle formed by the pages of an open book is:
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Acute
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Obtuse
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Right
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Straight
Explanation
The angle formed by the pages of an open book is obtuse.
Since the angle will always be more than $$90$$ but less than $$180$$.
How many pairs of adjacent angles, in all, can you name in figure?
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8
0%
7
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12
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10
Explanation
The pairs of adjacent angles are:
1. $$\angle AOB$$ and $$\angle BOC$$
2. $$\angle AOB$$ and $$\angle BOD$$
3. $$\angle AOB$$ and $$\angle BOE$$
4. $$\angle BOC$$ and $$\angle COD$$
5. $$\angle BOC$$ and $$\angle COE$$
6. $$\angle COD$$ and $$\angle AOC$$
7. $$\angle COD$$ and $$\angle DOE$$
8. $$\angle DOE$$ and $$\angle BOD$$
9. $$\angle DOE$$ and $$\angle AOD$$
10.
$$\angle COE$$ and $$\angle AOC$$
Hence there are 10 pairs of adjacent angles.
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