RS AGGARWAL CLASS 9 CHAPTER 7 LINES AND ANGLES EXERCISE 7B

 EXERCISE 7B

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Question 1:

In the adjoining figure, AOB is a straight line. Find the value of x.

Answer 1:

We know that the sum of angles in a linear pair is $180°$.
Therefore,
$$\begin{gathered} \angle AOC+\angle BOC=180° \\ \Rightarrow 62°+x°=180° \\ \Rightarrow x°=\left(180°-62°\right) \\ \Rightarrow x=118° \end{gathered}$$
Hence, the value of x is $118°$.

Question 2:

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC and ∠BOD.

Answer 2:

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is $180°$.
Therefore,
 $$\begin{gathered} \angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180° \\ \Rightarrow \left(3x-7\right)°+55°+\left(x+20\right)°=180 \\ \Rightarrow 4x=112° \\ \Rightarrow x=28° \end{gathered}$$
Hence,
$\angle \mathrm{AOC}=3x-7$
         $$\begin{gathered} =3\times 28-7 \\ =77° \end{gathered}$$
and $\angle \mathrm{BOD}=x+20$
                $$\begin{gathered} =28+20 \\ =48° \end{gathered}$$

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Question 3:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

Answer 3:

AOB is a straight line. Therefore,
$$\begin{gathered} \angle AOC+\angle COD+\angle BOD=180° \\ \Rightarrow \left(3x+7\right)°+\left(2x-19\right)°+x°=180° \\ \Rightarrow 6x=192° \\ \Rightarrow x=32° \end{gathered}$$
Therefore,
$$\begin{gathered} \angle AOC=3\times 32°+7=103° \\ \angle COD=2\times 32°-19=45°\mathbf{ }\mathrm{and} \\ \angle BOD=32° \end{gathered}$$

Question 4:

In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.

Answer 4:

Let $x=5a, y=4a \text{and} z=6a$
XOY is a straight line. Therefore,

$$\begin{gathered} \angle XOP+\angle POQ+\angle YOQ=180° \\ \Rightarrow 5a+4a+6a=180° \\ \Rightarrow 15a=180° \\ \Rightarrow a=12° \end{gathered}$$
Therefore,

$$\begin{gathered} x\Rightarrow 5\times 12°=60° \\ y\Rightarrow 4\times 12°=48° \\ \mathrm{and} z\Rightarrow 6\times 12°=72° \end{gathered}$$

Question 5:

In the adjoining figure, what value of x will make AOB a straight line?

Answer 5:

AOB will be a straight line if
$$\begin{gathered} 3x+20+4x-36=180° \\ \Rightarrow 7x=196° \\ \Rightarrow x=28° \end{gathered}$$
Hence, x = 28 will make AOB a straight line.

Question 6:

Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.

Answer 6:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore, 
$$\begin{gathered} \angle AOC+\angle AOD+\angle BOD+\angle BOC=360° \\ \Rightarrow 50+x+50+x=360° \\ \Rightarrow 2x=260° \\ \Rightarrow x=130° \end{gathered}$$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, $\angle AOD=130°, \angle BOD=50° \text{and} \angle BOC=130°$.

Question 7:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

Answer 7:

We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°$
Hence, $t=90°$
Also, 
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$$\begin{gathered} \angle AOC+\angle COE+\angle BOE=180° \\ \Rightarrow 90+50+y=180° \\ \Rightarrow 140+y=180° \\ \Rightarrow y=40° \end{gathered}$$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$
$\therefore x=40°, y=40°, z=50° \text{and} t=90°$

Question 8:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ∠AOD, ∠COE and ∠AOE.

Answer 8:

We know that if two lines intersect, then the vertically-opposite angles are equal.
$$\begin{gathered} \therefore \angle DOF=\angle COE=5x° \\ \angle AOD=\angle BOC=2x° \text{and} \\ \angle AOE=\angle BOF=3x° \end{gathered}$$

Since, AOB is a straight line, we have:
$$\begin{gathered} \angle AOE+\angle COE+\angle BOC=180° \\ \Rightarrow 3x+5x+2x=180° \\ \Rightarrow 10x=180° \\ \Rightarrow x=18° \end{gathered}$$

Therefore,
$$\begin{gathered} \angle AOD=2\times 18°=36° \\ \angle COE=5\times 18°=90° \\ \angle AOE=3\times 18°=54° \end{gathered}$$

Question 9:

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.

Answer 9:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
$$\begin{gathered} 5x+4x=180° \\ \Rightarrow 9x=180° \\ \Rightarrow x=20° \end{gathered}$$
Hence, the two angles are $5\times 20°=100° \text{and} 4\times 20°=80°$.

Question 10:

If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.

Answer 10:

We know that if two lines intersect, then the vertically-opposite angles are equal.

$\angle AOC=90°. \text{T}\mathrm{hen}, \angle AOC=\angle BOD=90°.$
And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$$\begin{gathered} \therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360° \\ \Rightarrow 90°+90°+x+x=360° \\ \Rightarrow 2x=180° \\ \Rightarrow x=90° \end{gathered}$$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90^o.

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Question 3:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

Answer 3:

AOB is a straight line. Therefore,
$$\begin{gathered} \angle AOC+\angle COD+\angle BOD=180° \\ \Rightarrow \left(3x+7\right)°+\left(2x-19\right)°+x°=180° \\ \Rightarrow 6x=192° \\ \Rightarrow x=32° \end{gathered}$$
Therefore,
$$\begin{gathered} \angle AOC=3\times 32°+7=103° \\ \angle COD=2\times 32°-19=45°\mathbf{ }\mathrm{and} \\ \angle BOD=32° \end{gathered}$$

Question 4:

In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.

Answer 4:

Let $x=5a, y=4a \text{and} z=6a$
XOY is a straight line. Therefore,

$$\begin{gathered} \angle XOP+\angle POQ+\angle YOQ=180° \\ \Rightarrow 5a+4a+6a=180° \\ \Rightarrow 15a=180° \\ \Rightarrow a=12° \end{gathered}$$
Therefore,

$$\begin{gathered} x\Rightarrow 5\times 12°=60° \\ y\Rightarrow 4\times 12°=48° \\ \mathrm{and} z\Rightarrow 6\times 12°=72° \end{gathered}$$

Question 5:

In the adjoining figure, what value of x will make AOB a straight line?

Answer 5:

AOB will be a straight line if
$$\begin{gathered} 3x+20+4x-36=180° \\ \Rightarrow 7x=196° \\ \Rightarrow x=28° \end{gathered}$$
Hence, x = 28 will make AOB a straight line.

Question 6:

Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.

Answer 6:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore, 
$$\begin{gathered} \angle AOC+\angle AOD+\angle BOD+\angle BOC=360° \\ \Rightarrow 50+x+50+x=360° \\ \Rightarrow 2x=260° \\ \Rightarrow x=130° \end{gathered}$$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, $\angle AOD=130°, \angle BOD=50° \text{and} \angle BOC=130°$.

Question 7:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

Answer 7:

We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°$
Hence, $t=90°$
Also, 
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$$\begin{gathered} \angle AOC+\angle COE+\angle BOE=180° \\ \Rightarrow 90+50+y=180° \\ \Rightarrow 140+y=180° \\ \Rightarrow y=40° \end{gathered}$$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$
$\therefore x=40°, y=40°, z=50° \text{and} t=90°$

Question 8:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ∠AOD, ∠COE and ∠AOE.

Answer 8:

We know that if two lines intersect, then the vertically-opposite angles are equal.
$$\begin{gathered} \therefore \angle DOF=\angle COE=5x° \\ \angle AOD=\angle BOC=2x° \text{and} \\ \angle AOE=\angle BOF=3x° \end{gathered}$$

Since, AOB is a straight line, we have:
$$\begin{gathered} \angle AOE+\angle COE+\angle BOC=180° \\ \Rightarrow 3x+5x+2x=180° \\ \Rightarrow 10x=180° \\ \Rightarrow x=18° \end{gathered}$$

Therefore,
$$\begin{gathered} \angle AOD=2\times 18°=36° \\ \angle COE=5\times 18°=90° \\ \angle AOE=3\times 18°=54° \end{gathered}$$

Question 9:

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.

Answer 9:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
$$\begin{gathered} 5x+4x=180° \\ \Rightarrow 9x=180° \\ \Rightarrow x=20° \end{gathered}$$
Hence, the two angles are $5\times 20°=100° \text{and} 4\times 20°=80°$.

Question 10:

If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.

Answer 10:

We know that if two lines intersect, then the vertically-opposite angles are equal.

$\angle AOC=90°. \text{T}\mathrm{hen}, \angle AOC=\angle BOD=90°.$
And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$$\begin{gathered} \therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360° \\ \Rightarrow 90°+90°+x+x=360° \\ \Rightarrow 2x=180° \\ \Rightarrow x=90° \end{gathered}$$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90^o.