Lines and Angles — NCERT Solutions, AI Tutor & Practice

When two lines intersect, vertically opposite angles are equal.

• $\angle BOD$ is vertically opposite to $\angle AOC$, so $\angle BOD = 50^\circ$.

Now $\angle AOC$ and $\angle AOD$ form a linear pair, so they add to $180^\circ$: $\angle AOD = 180^\circ - 50^\circ = 130^\circ$

• $\angle BOC$ is vertically opposite to $\angle AOD$, so $\angle BOC = 130^\circ$.

Answer: $\angle BOD = 50^\circ$, $\angle AOD = 130^\circ$, $\angle BOC = 130^\circ$.

Since ray $OP$ stands on the straight line $LM$, $\angle LOP$ and $\angle MOP$ form a linear pair: $\angle LOP + \angle MOP = 180^\circ$ $(2x+15) + (3x-25) = 180$ $5x - 10 = 180$ $5x = 190 \implies x = 38$

Now substitute:

• $\angle LOP = 2(38) + 15 = 91^\circ$
• $\angle MOP = 3(38) - 25 = 89^\circ$

Check: $91^\circ + 89^\circ = 180^\circ$ ✓

Let the two adjacent angles be $4k$ and $5k$. Since they lie on a straight line (linear pair): $4k + 5k = 180^\circ$ $9k = 180^\circ \implies k = 20^\circ$

Therefore:

• First angle $= 4k = 80^\circ$
• Second angle $= 5k = 100^\circ$

Answer: $80^\circ$ and $100^\circ$.

The angles around a point add up to a complete rotation: $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 = 360^\circ$

Since three lines cross at $O$, angles occur in vertically opposite pairs:

• $\angle 1$ and its opposite are equal $= 40^\circ$
• $\angle 2$ and its opposite are equal $= 65^\circ$

The third pair of opposite angles: on one side of a line, $\angle 1 + \angle 2 + \text{third angle} = 180^\circ$ (angles on a straight line). $\text{third angle} = 180^\circ - 40^\circ - 65^\circ = 75^\circ$

So the six angles are $40^\circ, 65^\circ, 75^\circ, 40^\circ, 65^\circ, 75^\circ$.

Sum $= 2(40^\circ + 65^\circ + 75^\circ) = 2 \times 180^\circ = 360^\circ$ ✓

Hence the vertically opposite pairs are consistent and the total is $360^\circ$.

Given $l \parallel m$ with transversal $t$, and one angle $= 110^\circ$.

Corresponding angle: Corresponding angles are equal, so the corresponding angle $= 110^\circ$.

Alternate interior angle: Alternate interior angles are equal, so it $= 110^\circ$.

Co-interior (allied) angle: Co-interior angles are supplementary: $180^\circ - 110^\circ = 70^\circ$

Answer: corresponding $= 110^\circ$, alternate interior $= 110^\circ$, co-interior $= 70^\circ$.

Since $AB \parallel CD$, co-interior (same-side interior) angles are supplementary: $(3x - 10) + (2x + 40) = 180$ $5x + 30 = 180$ $5x = 150 \implies x = 30$

Now:

• First angle $= 3(30) - 10 = 80^\circ$
• Second angle $= 2(30) + 40 = 100^\circ$

Check: $80^\circ + 100^\circ = 180^\circ$ ✓

Since $AB \parallel CD$, alternate interior angles are equal: $\angle APQ = \angle PQD$ $75 = 2y \implies y = 37.5$

So $\angle PQD = 75^\circ$.

Now $\angle PQC$ and $\angle PQD$ form a linear pair on line $CD$: $\angle PQC = 180^\circ - 75^\circ = 105^\circ$

Answer: $y = 37.5$, and $\angle PQC = 105^\circ$.

Since $CD$ passes through $C$ and $AB \parallel CD$: $\angle BCD = \angle ABC = 65^\circ \quad (\text{alternate interior angles})$

Since $CD \parallel EF$, angles $\angle DCE$ and $\angle CEF$ are co-interior: $\angle DCE = 180^\circ - 130^\circ = 50^\circ$

Therefore: $\angle BCE = \angle BCD + \angle DCE = 65^\circ + 50^\circ = 115^\circ$

Answer: $\angle BCE = 115^\circ$.

By the angle-sum property of a triangle: $\angle A + \angle B + \angle C = 180^\circ$ $65^\circ + 45^\circ + \angle C = 180^\circ$ $\angle C = 180^\circ - 110^\circ = 70^\circ$

The exterior angle at $C$ forms a linear pair with $\angle C$: $\text{Exterior angle} = 180^\circ - 70^\circ = 110^\circ$

(Also equal to $\angle A + \angle B = 65^\circ + 45^\circ = 110^\circ$, confirming the exterior angle theorem.)

By the exterior angle theorem, an exterior angle equals the sum of the two interior opposite angles: $120^\circ = 50^\circ + \text{(other opposite angle)}$ $\text{other opposite angle} = 120^\circ - 50^\circ = 70^\circ$

The third angle of the triangle is adjacent to the exterior angle (linear pair): $\text{third angle} = 180^\circ - 120^\circ = 60^\circ$

Check: $50^\circ + 70^\circ + 60^\circ = 180^\circ$ ✓

Answer: other interior opposite angle $= 70^\circ$, third angle $= 60^\circ$.

Let the angles be $2k$, $3k$, $4k$. By the angle-sum property: $2k + 3k + 4k = 180^\circ$ $9k = 180^\circ \implies k = 20^\circ$

Therefore the angles are:

• $2k = 40^\circ$
• $3k = 60^\circ$
• $4k = 80^\circ$

Since all angles are less than $90^\circ$, the triangle is an acute-angled triangle.

Answer: $40^\circ, 60^\circ, 80^\circ$ (acute-angled).

The exterior angle $\angle PRS$ (formed by producing $QR$) equals the sum of the two interior opposite angles $\angle P$ and $\angle Q$: $\angle PRS = \angle P + \angle Q$ $105^\circ = 60^\circ + \angle Q$ $\angle Q = 45^\circ$

Now use the angle-sum property to find $\angle R$: $\angle P + \angle Q + \angle R = 180^\circ$ $60^\circ + 45^\circ + \angle R = 180^\circ$ $\angle R = 75^\circ$

Check (exterior angle at R): $\angle PRS = 180^\circ - \angle R = 180^\circ - 75^\circ = 105^\circ$ ✓

Answer: $\angle Q = 45^\circ$, $\angle R = 75^\circ$.