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 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 $$AB$$.

A
$$66^o$$

B
$$46^o$$

C
$$60^o$$

D
none of the above

Solution

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.
Question 2
1 / -0

In the figure, given AE // BD and AC // ED: find $$\angle b$$.

A
$$65^o$$

B
$$36^o$$

C
$$45^o$$

D
none of the above

Solution

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.
Question 3
1 / -0

In the given figure, $$AB || CD$$, $$EH || BC$$, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of $$\angle CFG$$.

A
$$60^o$$

B
$$80^o$$

C
$$56^o$$

D
none of the above
Question 4
1 / -0

Two lines that are respectively parallel to two intersecting lines, intersect each other.

A
True

B
False

C
Ambiguous

D
Data Insufficient

Solution
Question 5
1 / -0

Two supplementary angles differ by $$48^o$$. Then find these angles.

A
$$36^o, 84^o$$

B
$$46^o, 94^o$$

C
$$56^o, 104^o$$

D
$$66^o, 114^o$$

Solution

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 $$66^o$$.
Question 6
1 / -0

Angles forming a linear pair can both be acute angles.

A
True

B
False

C
Ambiguous

D
Data Insufficient

Solution

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.
Question 7
1 / -0

Find the angle which is $$80^o$$ more than its complement.

A
$$75$$

B
$$85$$

C
$$95$$

D
$$100$$

Solution

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.
Question 8
1 / -0

How many pairs of adjacent angles, in all, can you name in figure?

A
8

B
7

C
12

D
10

Solution

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.
Question 9
1 / -0

If two adjacent angles are equal, then each angle measures $$90^o$$.

A
True

B
False

C
Ambiguous

D
Data Insufficient

Solution

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.
Question 10
1 / -0

Find the angle which is $$60^o$$ more than its complement.

A
$$55$$

B
$$75$$

C
$$85$$

D
$$60$$

Solution

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.