CBSE Class 9 Maths Chapter 6 Lines and Angles NCERT Solutions

Solution:

We are given that lines AB and CD intersect at O. The given information is:

• $\angle AOC + \angle BOE = 70^{\circ}$ (Equation 1)
• $\angle BOD = 40^{\circ}$ (Equation 2)

Since lines AB and CD intersect at O, the vertically opposite angles are equal. Therefore,

$\angle AOC = \angle BOD$

From Equation 2, we know $\angle BOD = 40^{\circ}$. So,

$\angle AOC = 40^{\circ}$

Now, substitute the value of $\angle AOC$ into Equation 1:

$40^{\circ} + \angle BOE = 70^{\circ}$

Subtracting $40^{\circ}$ from both sides, we get:

$\angle BOE = 70^{\circ} - 40^{\circ} = 30^{\circ}$

To find the reflex angle $\angle COE$, we first find the angle $\angle COE$. Since AOB is a straight line, the angles on this line sum up to $180^{\circ}$.

$\angle AOC + \angle COE + \angle BOE = 180^{\circ}$

We know $\angle AOC = 40^{\circ}$ and $\angle BOE = 30^{\circ}$. Substituting these values:

$40^{\circ} + \angle COE + 30^{\circ} = 180^{\circ}$

$\angle COE + 70^{\circ} = 180^{\circ}$

$\angle COE = 180^{\circ} - 70^{\circ} = 110^{\circ}$

The reflex angle $\angle COE$ is $360^{\circ}$ minus the angle $\angle COE$.

Reflex $\angle COE = 360^{\circ} - \angle COE = 360^{\circ} - 110^{\circ} = 250^{\circ}$

Therefore, $\angle BOE = 30^{\circ}$ and reflex $\angle COE = 250^{\circ}$.

Solution:

We are given that lines XY and MN intersect at O. We are also given:

• $\angle POY = 90^{\circ}$
• $a:b = 2:3$

Let $a = 2x$ and $b = 3x$ for some positive value $x$. The angles $\angle XOM$, $\angle POM$, and $\angle POY$ form a linear pair along the line XY.

Therefore, $\angle XOM + \angle POM + \angle POY = 180^{\circ}$

Substituting the given values and the expressions for $a$ and $b$:

$b + a + 90^{\circ} = 180^{\circ}$

$3x + 2x + 90^{\circ} = 180^{\circ}$

$5x = 180^{\circ} - 90^{\circ}$

$5x = 90^{\circ}$

$x = \frac{90^{\circ}}{5} = 18^{\circ}$

Now we can find the values of $a$ and $b$:

$a = 2x = 2 \times 18^{\circ} = 36^{\circ}$

$b = 3x = 3 \times 18^{\circ} = 54^{\circ}$

Angle $c$ is vertically opposite to $\angle XOM$. Also, $\angle XON$ is vertically opposite to $\angle MOY$. We need to find $c$, which is $\angle XON$.

Since MN is a straight line, $\angle XOM + \angle XON = 180^{\circ}$ (Linear pair).

We know $\angle XOM = b = 54^{\circ}$.

$54^{\circ} + c = 180^{\circ}$

$c = 180^{\circ} - 54^{\circ} = 126^{\circ}$

Alternatively, angle $c$ ($\angle XON$) is vertically opposite to $\angle MOY$. $\angle MOY = \angle MOP + \angle POY$. We found $\angle MOP = a = 36^{\circ}$ and $\angle POY = 90^{\circ}$.

$\angle MOY = 36^{\circ} + 90^{\circ} = 126^{\circ}$.

Therefore, $c = \angle XON = \angle MOY = 126^{\circ}$.

Hence, $c = 126^{\circ}$.

Solution:

We are given that $\angle PQR = \angle PRQ$.

Consider the line SRT. The angles $\angle PQS$ and $\angle PQR$ form a linear pair because they are adjacent angles on the straight line PT.

Therefore, $\angle PQS + \angle PQR = 180^{\circ}$ (Equation 1) (Linear pair axiom)

Similarly, consider the line PQT. The angles $\angle PRQ$ and $\angle PRT$ form a linear pair because they are adjacent angles on the straight line PT.

Therefore, $\angle PRQ + \angle PRT = 180^{\circ}$ (Equation 2) (Linear pair axiom)

From Equation 1 and Equation 2, we can see that the sum of angles in both cases is $180^{\circ}$.

$\angle PQS + \angle PQR = \angle PRQ + \angle PRT$

We are given that $\angle PQR = \angle PRQ$. Substituting this into the equation:

$\angle PQS + \angle PQR = \angle PQR + \angle PRT$

Subtracting $\angle PQR$ from both sides, we get:

$\angle PQS = \angle PRT$

Thus, it is proved that $\angle PQS = \angle PRT$.

Solution:

We are given the condition $x + y = w + z$.

We know that the sum of all angles around a point is $360^{\circ}$. Therefore,

$x + y + w + z = 360^{\circ}$

Since we are given $x + y = w + z$, we can substitute $(x + y)$ for $(w + z)$ in the equation:

$x + y + (x + y) = 360^{\circ}$

$2(x + y) = 360^{\circ}$

$x + y = \frac{360^{\circ}}{2}$

$x + y = 180^{\circ}$

The angles $x$ and $y$ are adjacent angles that form the angle $\angle AOB$. Since their sum is $180^{\circ}$, they form a linear pair.

According to the converse of the linear pair axiom, if the sum of two adjacent angles is $180^{\circ}$, then the non-common arms of the angles form a straight line.

In this case, the non-common arms are OA and OB. Therefore, AOB is a straight line.

Hence, it is proved that AOB is a line.

Solution:

We are given that POQ is a line and OR is perpendicular to PQ. This means that $\angle ROQ = 90^{\circ}$ and $\angle ROP = 90^{\circ}$.

We are also given that OS is a ray lying between rays OP and OR.

We need to prove that $\angle ROS = \frac{1}{2} (\angle QOS - \angle POS)$.

Consider the angle $\angle QOS$. Since OS lies between OR and OQ, we can write:

$\angle QOS = \angle QOR + \angle ROS$

We know $\angle QOR = 90^{\circ}$. So,

$\angle QOS = 90^{\circ} + \angle ROS$

Rearranging this equation to express $\angle ROS$:

$\angle ROS = \angle QOS - 90^{\circ}$ (Equation 1)

Now consider the angle $\angle POS$. Since OS lies between OP and OR, we can write:

$\angle ROP = \angle POS + \angle ROS$

We know $\angle ROP = 90^{\circ}$. So,

$90^{\circ} = \angle POS + \angle ROS$

Rearranging this equation to express $\angle ROS$:

$\angle ROS = 90^{\circ} - \angle POS$ (Equation 2)

Now we have two expressions for $\angle ROS$. Let's subtract Equation 2 from Equation 1:

$\angle ROS - \angle ROS = (\angle QOS - 90^{\circ}) - (90^{\circ} - \angle POS)$

$0 = \angle QOS - 90^{\circ} - 90^{\circ} + \angle POS$

$0 = \angle QOS + \angle POS - 180^{\circ}$

This approach doesn't seem to lead directly to the required proof. Let's try another method by manipulating the equations for $\angle QOS$ and $\angle POS$ differently.

From Equation 1: $\angle QOS = 90^{\circ} + \angle ROS$

From Equation 2: $\angle POS = 90^{\circ} - \angle ROS$

Now, let's find the difference $\angle QOS - \angle POS$:

$\angle QOS - \angle POS = (90^{\circ} + \angle ROS) - (90^{\circ} - \angle ROS)$

$\angle QOS - \angle POS = 90^{\circ} + \angle ROS - 90^{\circ} + \angle ROS$

$\angle QOS - \angle POS = 2 \angle ROS$

Now, divide both sides by 2:

$\frac{\angle QOS - \angle POS}{2} = \angle ROS$

Therefore, $\angle ROS = \frac{1}{2} (\angle QOS - \angle POS)$.

This proves the required statement.