Exercise 6.2: Lines and Angles — NCERT Solutions

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