CBSE Class 9 Maths Chapter 6 Lines and Angles NCERT Solutions

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

$$\angle AOC\; +\; \angle BOE\; =\; 70°$$ (1)

$$\angle BOD\; =\; 40°$$ (2)

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

$$\angle AOC\; =\; \angle BOD$$ (Vertically opposite angles)

From equation (2), we know $\angle BOD\; =\; 40°$. So,

$$\angle AOC\; =\; 40°$$

Now, substitute the value of $\angle AOC$ into equation (1):

$$40°\; +\; \angle BOE\; =\; 70°$$

Solving for $\angle BOE$:

$$\angle BOE\; =\; 70°\; -\; 40°\; =\; 30°$$

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 to 180 degrees:

$$\angle AOC\; +\; \angle COE\; +\; \angle BOE\; =\; 180°$$ (Angles on a straight line)

We know $\angle AOC\; +\; \angle BOE\; =\; 70°$ from equation (1). Substituting this:

$$70°\; +\; \angle COE\; =\; 180°$$

Solving for $\angle COE$:

$$\angle COE\; =\; 180°\; -\; 70°\; =\; 110°$$

The reflex angle $\angle COE$ is calculated as:

$$Reflex\; \angle COE\; =\; 360°\; -\; \angle COE\; =\; 360°\; -\; 110°\; =\; 250°$$

Therefore, $\angle BOE\; =\; 30°$ and reflex $\angle COE\; =\; 250°$.

We are given that lines XY and MN intersect at O, and $\angle POY\; =\; 90°$. We are also given the ratio a:b = 2:3.

Let $a\; =\; 2x$ and $b\; =\; 3x$, where x is a common factor.

Angles $\angle XOM$ (which is angle b) and $\angle POM$ (which is angle a) are adjacent angles on the straight line XY. Therefore, they form a linear pair with $\angle POY$.

The sum of angles on a straight line is 180 degrees. Thus,

$$\angle XOM\; +\; \angle POM\; +\; \angle POY\; =\; 180°$$ (Linear pair axiom)

Substituting the values and expressions:

$$3x\; +\; 2x\; +\; 90°\; =\; 180°$$

Combine the terms with x:

$$5x\; =\; 180°\; -\; 90°$$

$$5x\; =\; 90°$$

Solve for x:

$$x\; =\; \backslash frac\{90°\}\{5\}\; =\; 18°$$

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

$$a\; =\; \angle POM\; =\; 2x\; =\; 2\; \backslash times\; 18°\; =\; 36°$$

$$b\; =\; \angle XOM\; =\; 3x\; =\; 3\; \backslash times\; 18°\; =\; 54°$$

Angle c ($\angle XON$) is vertically opposite to angle $\angle MOY$. The angle $\angle MOY$ is the sum of $\angle MOP$ (angle a) and $\angle POY$.

$$\angle MOY\; =\; \angle MOP\; +\; \angle POY\; =\; a\; +\; 90°\; =\; 36°\; +\; 90°\; =\; 126°$$

Since angle c is vertically opposite to $\angle MOY$:

$$c\; =\; \angle XON\; =\; \angle MOY\; =\; 126°$$ (Vertically opposite angles)

Therefore, $c\; =\; 126°$.

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

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

$$\angle PQS\; +\; \angle PQR\; =\; 180°$$ (1) (Linear pair axiom)

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

$$\angle PRQ\; +\; \angle PRT\; =\; 180°$$ (2) (Linear pair axiom)

From equation (1), we can express $\angle PQS$ as:

$$\angle PQS\; =\; 180°\; -\; \angle PQR$$

From equation (2), we can express $\angle PRT$ as:

$$\angle PRT\; =\; 180°\; -\; \angle PRQ$$

We are given that $\angle PQR\; =\; \angle PRQ$. Substituting this into the expressions for $\angle PQS$ and $\angle PRT$:

$$\angle PQS\; =\; 180°\; -\; \angle PQR$$

$$\angle PRT\; =\; 180°\; -\; \angle PQR$$ (since $\angle PRQ\; =\; \angle PQR$)

Therefore, we can conclude that:

$$\angle PQS\; =\; \angle PRT$$

Proved.

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

We know that the sum of all angles around a point is 360 degrees. Therefore,

$$x\; +\; y\; +\; w\; +\; z\; =\; 360°$$

Since we are given that $x\; +\; y\; =\; w\; +\; z$, we can substitute $x\; +\; y$ for $w\; +\; z$ in the equation above:

$$(x\; +\; y)\; +\; (x\; +\; y)\; =\; 360°$$

$$2(x\; +\; y)\; =\; 360°$$

Divide by 2:

$$x\; +\; y\; =\; 180°$$

The angles x and y are adjacent angles that form the angle $\angle AOB$. Since their sum is 180 degrees, they form a linear pair. According to the converse of the linear pair axiom, if the sum of two adjacent angles is 180 degrees, 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.

Proved.

We are given that POQ is a line and OR is perpendicular to PQ. This means that $\angle ROP\; =\; 90°$ and $\angle ROQ\; =\; 90°$.

OS is a ray lying between OP and OR.

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

Let's express $\angle ROS$ in terms of other angles:

Since OS lies between OP and OR, we can write:

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

We know $\angle ROP\; =\; 90°$. So,

$$90°\; =\; \angle ROS\; +\; \angle POS$$

From this, we can express $\angle ROS$ as:

$$\angle ROS\; =\; 90°\; -\; \angle POS$$ (1)

Now, let's consider the angle $\angle QOS$. Since OS lies between OR and OQ, we can write:

$$\angle ROQ\; =\; \angle ROS\; +\; \angle QOS$$

Wait, this is incorrect. OS lies between OP and OR. So, $\angle QOS$ can be expressed using $\angle ROQ$ and $\angle ROS$.

We know $\angle ROQ\; =\; 90°$. Also, $\angle QOS$ is the sum of $\angle ROQ$ and $\angle ROS$ if S was between R and Q. However, S is between P and R.

Let's re-evaluate $\angle QOS$. Since POQ is a straight line, $\angle QOS\; =\; \angle QOR\; +\; \angle ROS$.

We know $\angle QOR\; =\; 90°$.

$$\angle QOS\; =\; 90°\; +\; \angle ROS$$

Now, let's consider the term $\angle QOS\; -\; \angle POS$:

$$\angle QOS\; -\; \angle POS\; =\; (90°\; +\; \angle ROS)\; -\; \angle POS$$

From equation (1), we have $\angle POS\; =\; 90°\; -\; \angle ROS$. Substitute this into the expression:

$$\angle QOS\; -\; \angle POS\; =\; (90°\; +\; \angle ROS)\; -\; (90°\; -\; \angle ROS)$$

$$\angle QOS\; -\; \angle POS\; =\; 90°\; +\; \angle ROS\; -\; 90°\; +\; \angle ROS$$

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

Rearranging this equation to solve for $\angle ROS$:

$$\angle ROS\; =\; \backslash frac\{1\}\{2\}\; (\angle QOS\; -\; \angle POS)$$

Proved.