Lines and Angles Test

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Lines and Angles Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

In figure, $$PQ$$ is a mirror, $$AB$$ is the incident ray and $$BC$$ is the reflected ray. If $$\angle ABC=46^o$$, then $$\angle ABP$$ is equal to

A
$$44^o$$

B
$$67^o$$

C
$$13^o$$

D
$$62^o$$

Solution

Given
$$\angle ABC=46^o$$

We know that in a plane mirror,
the angle of the incident $$=$$ angle of reflection.
$$\Rightarrow \angle CBQ =\angle ABP\quad...(i)$$

Since, $$PQ$$ is straight line
$$\therefore \angle ABP +\angle ABC+\angle CBQ =180^o$$
$$\Rightarrow \angle ABP +\angle ABC +\angle ABP =180^o\qquad$$[from $$(i)$$]
$$\Rightarrow 2\angle ABP +46^o =180^o$$
$$\Rightarrow 2\angle ABP =180^o -46^o =134$$
$$\Rightarrow \angle ABP =\dfrac{134^o}{2}$$
$$\Rightarrow \angle ABP =67^o$$

Hence, option (B) is correct.
Question 2
1 / -0

The angles $$x-10^o$$ and $$190^o-x$$ are

A
interior angles on the same side of the transversal

B
making a linear pair

C
complementary

D
supplementary

Solution

Option (d) is correct.
Given that $$x-10^o+190^o-x$$
$$\Rightarrow 190^o -10^o$$
$$\Rightarrow180^o$$
Therefore, given angles are supplementary.
Question 3
1 / -0

In figure, $$POQ$$ is a line. If $$x=30^o$$, then $$\angle QOR$$ is

A
$$90^o$$

B
$$30^o$$

C
$$150^o$$

D
$$60^o$$

Solution

Option (a) is correct.
It has been given that $$POQ$$ is a straight line.
$$\therefore x+2y+3y=180^o$$
$$\Rightarrow 5y+30^o =180^o$$
$$\Rightarrow 5y=180^o -30^o$$
$$\Rightarrow y=\dfrac{150^o}{5}$$
$$\Rightarrow y=30^o$$
$$\because \angle QOR =3y$$
$$\therefore QOR=3\times 30^o$$
$$\Rightarrow \therefore \angle QOR=90^o$$
Question 4
1 / -0

If two supplementary angles are in the ratio $$1:2$$, then the bigger angle is

A
$$120^o$$

B
$$125^o$$

C
$$110^o$$

D
$$90^o$$

Solution

Option $$(A)$$ is correct.
Two supplementary angles are in the ratio $$1:2$$
So, let two angles be $$x$$ and $$2x$$
Given that, angles are supplementary
$$\therefore x+2x=180^o$$
$$\Rightarrow 3x=180$$
$$\Rightarrow x=\dfrac{180^o}{3}$$
$$\Rightarrow x=60^o$$
Therefore, bigger angle is $$2x=120^{\circ}.$$
Question 5
1 / -0

In figure, $$\angle AOC$$ and $$\angle BOC$$ form a pair of

A
vertically opposite angles.

B
complementary angles.

C
alternate interior angles.

D
linear pair of angles.

Solution

In figure, $$\angle AOC$$ and $$\angle BOC$$ lie on a straight line and on the same side of the straight line.
So, they form a linear pair of supplementary angles that is $$\angle AOC+\angle BOC=180^{\circ}$$.

Hence, option $$(D)$$ is correct.
Question 6
1 / -0

In which of the following figures, $$a$$ and $$b$$ are forming a pair of adjacent angles?

A

B

C

D

Solution

For adjacent angles both angles must have one vertex and one side common.
$$a$$ and $$b$$ have a common vertex and a common arm that's why they are an adjacent angles.
Hence, Option $$(D)$$ is correct.
Question 7
1 / -0

The difference of two complementary angles is $$30^o$$. Then, the angles are:

A
$$60^o, 30^o$$

B
$$70^o, 40^o$$

C
$$20^o, 50^o$$

D
$$105^o, 75^o$$

Solution

Let $$x$$ and $$y$$ be the angles.
Then, $$x+y=90^o ....(1)$$ [ Complementary angle]
Also, $$x-y=30^o ....(2)$$ [ Given ]
By addition of both equations, we get,
$$\Rightarrow 2x=120^o$$
$$\Rightarrow x=\dfrac{120^o}{2}$$
$$\Rightarrow x=60^o$$.
Putting the value of $$x$$ in equation $$(1)$$, we get,
$$\Rightarrow 60+y=90^o$$
$$\Rightarrow y=90^o -60^o$$
$$\Rightarrow y=30^o$$
Therefore, the required angles are $$30^o$$ and $$60^o$$.
Hence, option $$A$$ is correct.
Question 8
1 / -0

In figure, which one of the following is not true ?

A
$$\angle 1+\angle 5=180^o$$

B
$$\angle 2+\angle 5=180^o$$

C
$$\angle 3+\angle 8=180^o$$

D
$$\angle 2+\angle 3=180^o$$

Solution

Given, $$PQ$$ parallel to $$RS$$ and line $$l$$ is a transversal.
$$\therefore \angle 2+\angle 5=180^0$$ (1).....[ Co-interior angles=supplementary angles ]
$$\angle 3+\angle 8=180^0$$ (2) ..... [ Co- interior angles=supplementary angles ]
$$\angle 1=\angle 2$$ (3)..... [ Vertically opposite angles ]
$$\therefore \angle 1+\angle 5=180^o$$ [ By equation (1) and (3)]
$$\angle 2=\angle 3$$ [ Alternate interior angles ]
However, $$\angle 2 +\angle 3=180^o$$ cannot be true.
Hence, option $$D$$ is correct.
Question 9
1 / -0

In a pair of adjacent angles, (i) vertex is always common, (ii) one arm is always common, and (iii) uncommon arms are always opposite rays
Then

A
All (i), (ii) and (iii) are true

B
(iii) is flase

C
(i) is false but (ii) and (iii) are true

D
(ii) is false

Solution

Option $$(B)$$ is correct.
Adjacent angles have a common vertex and a common arm.
but uncommon arms are only opposite in linear pair.
So, they always do not need to be opposite.
So, statement $$(iii)$$ is false
Question 10
1 / -0

If an angle is $$60^o$$ less than two times of its supplement, then the greater angle is

A
$$100^o$$

B
$$80^o$$

C
$$60^o$$

D
$$120^o$$

Solution

Option (a) is correct.
Let an angle be $$x$$ and $$180^o-x$$ be supplementary angles.
So, according to question,
$$\Rightarrow x=2(180^o -x)-60^o$$
$$\Rightarrow x=360^o -2x-60^o$$
$$\Rightarrow x+2x=300^o$$
$$\Rightarrow 3x=300^o$$
$$\Rightarrow x=\dfrac{300^o}{3}$$
$$\Rightarrow x=100^o$$
$$\Rightarrow 180^0-x=80^0$$
Greater angle $$=100^0$$
Hence, option $$A$$ is correct