Exercise 6.3: Lines and Angles — NCERT Solutions

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