Lines and Angles Test

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Lines and Angles Test - 39

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Question 1
1 / -0

Angles which are both supplementary and vertically opposite are

A
$$95^o , 85^o$$

B
$$90^o, 90^o$$

C
$$100^o, 80^o$$

D
$$45^o, 45^o$$

Solution

As we know, vertically opposite angles are equal.
Let the angle be $$y$$
$$\Rightarrow y+y=180^o$$
$$\Rightarrow 2y=180^o$$
$$\Rightarrow y=90^o$$
Therefore, $$90^o$$ is asked angle for each.
Question 2
1 / -0

If the complement of an angle is $$79^o$$, then the angle will be of :

A
$$1^o$$

B
$$11^o$$

C
$$79^o$$

D
$$101^o$$

Solution

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

Given, measure of one complementary angle is $$79^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-79^o$$ $$=11^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 79^{o}=11^o$$.
Hence, option $$B$$ is correct.
Question 3
1 / -0

The angles $$x$$ and $$90^o -x$$ are:

A
supplementary

B
complementary

C
vertically opposite

D
making a linear pair

Solution

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

Sum of given angles = $$(90^o-x^o)+$$ $$x^o=90^o$$.
Hence, these are complementary angles.
Therefore, option $$B$$ is correct.
Question 4
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 5
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 6
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 7
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 8
1 / -0

In figure, lines $$l$$ and $$m$$ intersect each other at a point. Which of the following is false?

A
$$\angle a= \angle b$$

B
$$\angle d =\angle c$$

C
$$\angle a+ \angle d=180^o$$

D
$$\angle a=\angle d$$

Solution

The two lines $$l$$ and $$m$$ shown in the figure forms $$4$$ angles, $$a,\ b,\ c\ and\ d$$.
$$\angle a$$ and $$\angle b$$ are vertically opposite angles.
Hence, $$\angle a=\angle b$$.
Similarly, $$\angle c$$ and $$\angle d$$ are also vertically opposite angles.
Hence, $$\angle d=\angle c$$.
Also, $$\angle a$$ and $$\angle d$$ form a linear pair.
Hence, $$\angle a+\angle d=180^o$$

Since $$\angle a+\angle d=180^o$$, $$\angle a$$ cannot be equal to $$\angle d$$, unless they are both right angles, which is not the case here.

Hence, option D is false here.
Question 9
1 / -0

In figure, the value of $$x$$ is

A
$$110^o$$

B
$$46^o$$

C
$$64^o$$

D
$$150^o$$

Solution

Option (d) is correct.
As we knew, the angle made around a point is $$360$$ degree.
Then,
$$\therefore y+64^o +46^o +100^o -360^o$$
$$\Rightarrow y+210^o =360^o $$
$$\Rightarrow y=360^o -210^o$$
$$\Rightarrow y=150^o$$
Question 10
1 / -0

In figure, the value of $$y$$ is

A
$$30^o$$

B
$$15^o$$

C
$$20^o$$

D
$$22.5^o$$

Solution

Option (c) is correct.
All the angle lies on the straight line.
$$\therefore 6y+y+2y=180^o$$
$$\Rightarrow 9y=180^o$$
$$\Rightarrow y=\dfrac{180^o }{9}$$
$$\Rightarrow y=20^o$$