Lines and Angles Test

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Lines and Angles Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following pairs represents pairof corresponding angles.

A
$$\angle a$$ and $$\angle d$$

B
$$\angle a$$ and $$\angle e$$

C
$$\angle \mathrm { d }$$ and $$\angle i$$

D
$$\angle \mathrm { c }$$ and $$\angle \mathrm { h }$$

Solution
Question 2
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If the supplement of an angle is $$\dfrac { 5 }{ 2 } $$ times the complement of the same angle, find the supplementary angle.

A
$${ 150 }^{ \circ }$$

B
$${ 60 }^{ \circ }$$

C
$${ 30 }^{ \circ }$$

D
$${ 120 }^{ \circ }$$
Question 3
1 / -0

The complementary angle of $$75^{\circ}$$ is _______.

A
$$75^{\circ}$$

B
$$105^{\circ}$$

C
$$15^{\circ}$$

D
$$25^{\circ}$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$75^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-75^o$$ $$=15^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 75^{o}=15^o$$.
Hence, option $$C$$ is correct.
Question 4
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Complement of $$60^o$$ is?

A
$$30^o$$

B
$$130^o$$

C
$$120^o$$

D
$$90^o$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$60^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-60^o$$ $$=30^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 60^{o}=30^o$$.
Hence, option $$A$$ is correct.
Question 5
1 / -0

Mark the correct alternative of the following:
The sum of an angle and half of its complementary angle is $$75^o$$. The measure of the angle is?

A
$$40^o$$

B
$$50^o$$

C
$$60^o$$

D
$$80^o$$

Solution

Given, the sum of angle and half its complementary angle is $$75^o$$.
We know, sum of the complementary angles is $$90^{ \circ }$$.
Then, let the angles be $$x$$ and $$(90^o-x)$$.
Now, $$ x+\dfrac{1}{2}(90^o-x)={ 75 }^{ \circ }$$
$$\implies$$ $$ x-\dfrac{1}{2}x+45^o={ 75 }^{ \circ }$$
$$\implies$$ $$ \dfrac{1}{2}x={ 75 }^{ o }-45^o$$
$$\implies$$ $$ \dfrac{1}{2}x=30^o$$
$$\implies$$ $$x= 2\times30^o=60^o$$.

Hence, the measure of the required angle is $$60^o$$.

Therefore, option $$C$$ is correct.
Question 6
1 / -0

Mark the correct alternative of the following.
The sum of an angle and one third of its supplementary angle is $$90^o$$. The measure of the angle is?

A
$$135^o$$

B
$$120^o$$

C
$$60^o$$

D
$$45^o$$

Solution

Let, required angle be $$x$$ and its supplementary angle be $$(180-x)$$.
$$\therefore \angle x+\angle (180-x)=180^0$$

According to condition, the sum of an angle and one third of its supplementary angle is $$90^0$$
$$\angle x+\dfrac{1}3\angle (180-x)=90^0\\$$
$$\angle x+\dfrac{1}3(180^0-x)=90^0\\$$
$$\dfrac2{3}\angle x+60^0=90^0\\$$
$$\dfrac2{3}\angle x=30^0\\$$
$$\angle x=45^0\\$$
$$\therefore$$ measure of a required angle is $$45^0$$
Question 7
1 / -0

Mark the correct alternative of the following:
Two complementary angles are in the ratio $$2:3$$. The measure of the larger angle is?

A
$$60^o$$

B
$$54^o$$

C
$$66^o$$

D
$$48^o$$

Solution

Given, the complementary angles are in the ratio $$2:3$$.
Let the angles be $$2x$$ and $$3x$$.
We know, sum of the complementary angles is $$90^{ \circ }$$.
$$\Rightarrow 2x+3x={ 90 }^{ \circ }\\ \Rightarrow 5x=9{ 0 }^{ \circ }\\ \Rightarrow x=18^{ \circ }$$

Hence, the angles are:
$$2x=2\times 18^{ \circ }=36^{ \circ }$$
and $$ 3x=3\times 18^{ \circ }=54^{ \circ }$$.

Here, the larger angle is $$54^o$$.

Therefore, option $$B$$ is correct.
Question 8
1 / -0

Two supplementary angles are in the ratio $$3:2$$. The smaller angle measures?

A
$$108^o$$

B
$$81^o$$

C
$$72^o$$

D
$$68^o$$

Solution

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 $$180^o$$.
Then we have,
$$3x+2x=180^{o}$$
$$5x=180^o$$
or, $$x=36^o$$.
So the smaller angle is $$36^o\times 2=72^o$$.
Question 9
1 / -0

Mark the correct alternative of the following.
In figure, PQ$$||$$RS and $$\angle$$PAB$$=60^o$$ and $$\angle ACS=100^o$$. Then, $$\angle BAC=?$$

A
$$40^o$$

B
$$60^o$$

C
$$80^o$$

D
$$50^o$$

Solution

From the picture it is clear that $$\angle PAC=\angle ACS$$. [ Alternate angles]
So,
$$\angle PAB+\angle BAC=\angle ACS$$ [ Since $$\angle PAC=\angle PAB+\angle BAC$$]
or, $$\angle BAC=100^0-60^o$$
or, $$\angle BAC=40^o$$.
Question 10
1 / -0

An angle is its own complement. The measure of the angle is:

A
$$30^{o}$$

B
$$45^{o}$$

C
$$90^{o}$$

D
$$60^{o}$$

Solution

Let the measure of the required angle be $$x^{o}$$.
Since, the angle is its own complement, both the angle and its complement will be equal.
Then,
$$\implies x^{o}+x^{o}=90^{o}$$
$$\implies 2x=90^o$$
$$\implies x=\dfrac{90}{2}^o$$
$$\implies x=45^o$$.
Hence, the required angle measures $$45^{o}$$.
Therefore, option $$B$$ is correct.

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Re-Attempt Weekly Quiz Competition

Lines and Angles Test - 27

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Lines and Angles Test - 27

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SHARING IS CARING

Question 1

$$\angle a$$ and $$\angle d$$

$$\angle a$$ and $$\angle e$$

$$\angle \mathrm { d }$$ and $$\angle i$$

$$\angle \mathrm { c }$$ and $$\angle \mathrm { h }$$

Solution

Question 2

$${ 150 }^{ \circ }$$

$${ 60 }^{ \circ }$$

$${ 30 }^{ \circ }$$

$${ 120 }^{ \circ }$$

Question 3

$$75^{\circ}$$

$$105^{\circ}$$

$$15^{\circ}$$

$$25^{\circ}$$

Solution

Question 4

$$30^o$$

$$130^o$$

$$120^o$$

$$90^o$$

Solution

Question 5

$$40^o$$

$$50^o$$

$$60^o$$

$$80^o$$

Solution

Question 6

$$135^o$$

$$120^o$$

$$60^o$$

$$45^o$$

Solution

Question 7

$$60^o$$

$$54^o$$

$$66^o$$

$$48^o$$

Solution

Question 8

$$108^o$$

$$81^o$$

$$72^o$$

$$68^o$$

Solution

Question 9

$$40^o$$

$$60^o$$

$$80^o$$

$$50^o$$

Solution

Question 10

$$30^{o}$$

$$45^{o}$$

$$90^{o}$$

$$60^{o}$$

Solution

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