1). In figure, find the values of x and y and then show that AB II CD.
Solution: From figure,
x and 50⁰ are a linear pairs
∴ x + 50⁰ = 180⁰
or, x = 180 - 50 = 130⁰ ……(1)
Now, angles y and 130⁰ are vertically opposite angles,
∴ y = 130⁰ …….(2)
from (1) and (2)
x = y = 130⁰
∵ We see that x and y are alternate angle, these are equal [ x = y = 130⁰]
Hence, AB II CD proved.
2). In figure, if AB ll CD, CD II EF and y : z = 3:7, find x.
Solution, Given, y : z = 3 : 7
Let y = 3k and z = 7k
Now, AB II CD II EF
∴ AB II EF, then x = z [ alternate angle]
or, x = z = 7k
∵ AB II CD
∴ x + y = 180⁰ [ interior angle]
∴ 7k + 3k = 180
or, 10k = 180
or, k = 180/10 = 18⁰
Hence, x = 7k = 7 × 18 = 126⁰
3). In figure, if AB II CD, EF ⏊CD and ∠GED = 126⁰, find ∠AGE, ∠GEF and ∠FGE.
Solution: Given, ∠GED = 126⁰
∵ AB IICD and GE is a transversal lines,
then ∠GED = ∠AGE = 126⁰[ Alternate angles are equal]
∵ ∠GED = ∠GEF + ∠ FED = 126⁰
or, ∠GEF + 90⁰ = 126⁰ [ ∵ ∠FED = 90⁰]
or, ∠GEF = 126 - 90 = 36⁰
Now, ∠GEC + ∠ GED = 180 [ linear pair]
or, ∠GEC + 126 = 180
or, ∠GEC = 180 - 126 = 54⁰
But, ∠GEC = ∠FGE = 54⁰[ alternate angle]
Hence, ∠AGE = 126⁰, ∠GEF = 36⁰ and ∠FGE = 54⁰
4). In figure, PQ II ST, ∠PQR = 110⁰, and ∠RST = 130⁰, find ∠QRS.
Solution: Given,
PQ II ST, ∠PQR = 110⁰ , ∠RST = 130⁰
Construction : Draw a line ST ll AB, passing through point R, such that
PQ ll ST ll AB
Now, ST ll AB, SR is a transversal line
∴ ∠RST + ∠SRT = 180⁰ [ ∵ corresponding interior angles]
or, 130⁰ + ∠ SRB = 180
or, ∠SRB = 180 - 130 = 50⁰
Similarly, PQ ll AB, QR is a transversal line, then
∠PQR + ∠QRA = 180 [ ∵ corresponding interior angles]
or, 110 + ∠QRA = 180
or, ∠QRA = 180 - 110 = 70⁰
∵ ARB is a straight line.
∴ ∠QRA + ∠QRS + ∠SRB = 180
or, 70 + x + 50 = 180
or, 120 + x = 180
or, x = 180 - 120 = 60⁰
Hence, ∠QRS = 60⁰
5). In figure, if AB ll CD, ∠APQ = 50⁰ and ∠PRD = 127⁰, find x and y.
Solution: Given, ∠APQ = 50⁰, ∠PRD = 127⁰ and AB ll CD
Now, AB ll CD, PR is a transversal line.
∴ ∠APR = ∠PRD [ alternate angle]
or, 50⁰ + y = 127⁰
or, y = 127 - 50 = 77⁰
Also, ∠APQ = ∠PQR [ alternate angle]
or, 50⁰ = x
Hence, x = 50⁰ and y = 77⁰
6). In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. prove that AB ll CD
Solution: Draw BE and CF normals to the mirrors PQ and RS at B and C respectively. Then, BE ⏊ PQ and CF ⏊ RS
∴ BE ll CF
Now, BE ll CF and BC is a transversal line.
∴ ∠3 = ∠ 2 …..(1) [ alternate angle]
But, ∠3 = ∠4 and ∠1 = ∠2 [ ∵ angle of incidence = angle of reflection]
Substitute, ∠3 = ∠4 and ∠1 = ∠2 in (1)
then, ∠4 = ∠1 …….(2)
Adding (1) and (2)
∴ ∠3 + ∠4 = ∠1 + ∠2
or, ∠ABC = ∠BCD
∵ ∠ABC and ∠BCD are the alternate angles of AB ll CD and BC is a transversal line
∴ AB ll CD proved.
