Lines & Angles - Exercise 17(C)

Two lines AB and CD are cut by a transversal EF, as shown in the figure. Identify the given pair of angles as adjacent angles, vertically opposite angles, alternate angles, corresponding angles or co-interior angles.

(i) ∠6 and ∠7

(ii) ∠3 and ∠4

(iii) ∠4 and ∠8

(iv) ∠1 and ∠5

(v) ∠3 and ∠5

(vi) ∠2 and ∠4

(vii) ∠4 and ∠5

(viii) ∠2 and ∠7

(ix) ∠3 and ∠6

(x) ∠4 and ∠6

(xi) ∠2 and ∠6

(xii) ∠1 and ∠4

Answer

(i) ∠6 and ∠7

These angles are opposite each other at the same intersection point formed by two intersecting lines CD and EF.

∴ These are vertically opposite angles.

(ii) ∠3 and ∠4

These angles share a common vertex and a common arm on line AB, and lie next to each other without overlapping.

∴ These are adjacent angles.

(iii) ∠4 and ∠8

These angles are in the same relative position (bottom-right) at each intersection.

∴ These are corresponding angles.

(iv) ∠1 and ∠5

These angles are in the "top-left" position at each intersection.

∴ These are corresponding angles.

(v) ∠3 and ∠5

These angles lie inside the two parallel lines and are on the same side of the transversal.

∴ These are co-interior angles.

(vi) ∠2 and ∠4

These angles share a common vertex and a common arm and lie next to each other without overlapping.

∴ These are adjacent angles.

(vii) ∠4 and ∠5

These angles lie between the two parallel lines but on opposite sides of the transversal.

∴ These are interior alternate angles.

(viii) ∠2 and ∠7

These angles lie outside the two parallel lines and on opposite sides of the transversal.

∴ These are exterior alternate angles.

(ix) ∠3 and ∠6

These angles lie between the two parallel lines but on opposite sides of the transversal.

∴ These are interior alternate angles.

(x) ∠4 and ∠6

These angles lie inside the two parallel lines and on the same side of the transversal.

∴ These are co-interior angles.

(xi) ∠2 and ∠6

These angles are in the same relative position (top-right) at different intersections.

∴ These are corresponding angles.

(xii) ∠1 and ∠4

These angles are opposite each other at the same intersection point formed by two intersecting lines AB and EF.

∴ These are vertically opposite angles.

In the given figure, AB || CD and EF is a transversal. If ∠8 = 110°, find each one of the unknown angles, marked in the figure. Give reasons.

Answer

Given:

AB || CD

EF is a transversal

∠8 = 110°

Let's find angles at the bottom intersection (Point on CD):

∠5 = ∠8 $\quad$[Vertically opposite angles]

∴ ∠5 = 110°

Since CD is a straight line, the angles ∠7 and ∠8 form a linear pair and must sum to 180°.

∴ ∠7 + ∠8 = 180°

⇒ ∠7 + 110° = 180° $\quad$[Substituting the value of ∠8]

⇒ ∠7 = 180° - 110°

⇒ ∠7 = 70°

∠6 = ∠7 $\quad$[Vertically opposite angles]

∴ ∠6 = 70°

Let's find angles at the top intersection (Point on AB):

∠4 = ∠8 $\quad$[Corresponding angles]

∴ ∠4 = 110°

∠1 = ∠4 $\quad$[Vertically opposite angles]

∴ ∠1 = 110°

Since AB is a straight line, the angles ∠1 and ∠2 form a linear pair and must sum to 180°.

∴ ∠1 + ∠2 = 180°

⇒ 110° + ∠2 = 180° $\quad$[Substituting the value of ∠1]

⇒ ∠2 = 180° - 110°

⇒ ∠2 = 70°

∠3 = ∠2 $\quad$[Vertically opposite angles]

∴ ∠3 = 70°

∠5 = 110°, ∠6 = 70°, ∠7 = 70°, ∠1 = 110°, ∠2 = 70°, ∠3 = 70°, ∠4 = 110°

In each of the following figures, AB || CD and EF is a transversal. Find each one of the unknown angles x, y, z in each case.

(i)

(ii)

(iii)

Answer

Given:

AB || CD

EF is a transversal

(i)

x° = 55° $\quad$[Vertically opposite angles are equal]

y° and 55° angle are alternate interior angles i.e., these angles are on opposite sides of the transversal between the parallel lines. So, they are equal.

∴ y° = 55°

Since y° and z° form a linear pair on line CD they must sum to 180°.

∴ y° + z° = 180°

⇒ 55° + z° = 180° $\quad$[Substituting the value of y]

⇒ z° = 180° - 55°

⇒ z° = 125°

x° = 55°, y° = 55°, z° = 125°

(ii)

x° and 130° are corresponding angles i.e., these angles are in the same relative position at each intersection. So, they are equal.

∴ x° = 130°

Since x° and y° form a linear pair on line CD they must sum to 180°.

∴ x° + y° = 180°

⇒ 130° + y° = 180° $\quad$[Substituting the value of x]

⇒ y° = 180° - 130°

⇒ y° = 50°

z° = y° $\quad$[Corresponding angles]

∴ z° = 50°

x° = 130°, y° = 50°, z° = 50°

(iii)

From the figure,

z° = 40° $\quad$[Vertically opposite angles]

z° = y° $\quad$[Interior alternate angles]

∴ y° = 40°

Since x° and y° form a linear pair on line AB they must sum to 180°.

∴ x° + y° = 180°

⇒ x° + 40° = 180° $\quad$[Substituting the value of y]

⇒ x° = 180° - 40°

⇒ x° = 140°

x° = 140°, y° = 40°, z° = 40°

In each of the following figures, AB || CD and EF is a transversal. Find the value of x in each case.

(i)

(ii)

(iii)

Answer

(i)

From the figure,

The opposite angle of 5x° will be equal to 5x°. Because, they are vertically opposite angles.

Now, 5x° and 3x° are co-interior angles i.e., they lie inside the parallel lines on the same side of the transversal.

Co-interior angles are supplementary:

∴ 5x° + 3x° = 180°

⇒ 8x° = 180°

⇒ x° = $\dfrac{180^{\circ}}{8}$

x° = 22.5°

(ii)

From the figure,

The opposite angle of 5x° will be equal to 5x°. Because, they are vertically opposite angles.

5x° and 4x° are co-interior angles i.e., they lie inside the parallel lines on the same side of the transversal.

Co-interior angles are supplementary:

∴ 5x° + 4x° = 180°

⇒ 9x° = 180°

⇒ x° = $\dfrac{180^{\circ}}{9}$

x° = 20°

(iii)

From the figure,

140° and 4x° are corresponding angles i.e., they are in the same relative position at each intersection. Therefore, they are equal.

4x° = 140°

⇒ x° = $\dfrac{140^{\circ}}{4}$

x° = 35°

In the given figure, AB || CD. If ∠BAC = (3x + 15)° and ∠ACD = (2x + 45)°, find the value of x.

Also, find the measures of ∠BAC and ∠ACD.

Answer

Given:

AB || CD

∠BAC = (3x + 15)°

∠ACD = (2x + 45)°

∠BAC and ∠ACD lie inside the parallel lines on the same side of transversal AC. So, they are co-interior angles.

Co-interior angles are supplementary:

∴ (3x + 15)° + (2x + 45)° = 180°

⇒ 3x° + 2x° + 15° + 45° = 180°

⇒ 5x° + 60° = 180°

⇒ 5x° = 180° - 60°

⇒ 5x° = 120°

⇒ x° = $\dfrac{120^{\circ}}{5}$

⇒ x° = 24°

Let's find each angle by substituting the value of x:

∠BAC = (3x + 15)° = (3(24) + 15)° = (72 + 15)° = 87°

∠ACD = (2x + 45)° = (2(24) + 45)° = (48 + 45)° = 93°

x° = 24°, ∠BAC = 87° and ∠ACD = 93°

In the given figure, AB || CD. Find the value of x.

Answer

Draw a line EF through point M such that EF is parallel to both AB and CD.

From the figure,

∠PAB = 3x°, ∠OCD = 2x° and ∠PMO = 100°

For AB || ME:

∠AME = ∠PAB $\quad$[Corresponding angles]

∴ ∠AME = 3x° $\quad$[Substituting the value of ∠PAB]

For CD || ME:

∠CME = ∠OCD $\quad$[Corresponding angles]

∴ ∠CME = 2x° $\quad$[Substituting the value of ∠OCD]

As the line ME splits the 100° angle at point M into two parts:

∴ ∠AME + ∠CME = 100°

⇒ 3x° + 2x° = 100°

⇒ 5x° = 100°

⇒ x° = $\dfrac{100^{\circ}}{5}$

x° = 20°

In the given figure, AB || DC and BC || AD. Find the values of x, y and z.

Answer

From the figure,

∠ABC = 110°, ∠BCD = x°, ∠ADC = z°, ∠DCE = y°

Since AB || CD and BC || AD, the figure ABCD is a parallelogram.

Consider parallel lines AB and CD with BC acting as a transversal.

Angles ∠ABC and ∠BCD are co-interior angles.

Co-interior angles are supplementary:

∴ ∠ABC + ∠BCD = 180°

⇒ 110° + x° = 180° $\quad$ [Substituting the values of ∠ABC and ∠BCD]

⇒ x° = 180° - 110°

⇒ x° = 70°

Angles ∠BCD (x°) and ∠DCE (y°) form a linear pair. So, they must sum to 180°.

∴ ∠BCD + ∠DCE = 180°

⇒ 70° + y° = 180° $\quad$ [Substituting the values of ∠BCD and ∠DCE]

⇒ y° = 180° - 70°

⇒ y° = 110°

In a parallelogram, opposite angles are equal.

Angle z° is opposite to ∠ABC

∴ z° = ∠ABC

z° = 110° $\quad$ [Substituting the values of ∠ABC]

x° = 70°, y° = 110° and z° = 110°

In each of the figures given below, AB || CD. Find the unknown angles, giving reasons.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Answer

(i)

Given:

AB || CD

∠ECD = 75°

∠AEC = ∠ECD $\quad$[Alternate angles]

⇒ y° = ∠ECD

⇒ y° = 75° $\quad$[Substituting the value of ∠ECD]

Since, AB is a straight line:

∠BEC + ∠AEC = 180° $\quad$[Linear pair]

⇒ x° + y° = 180°

⇒ x° + 75° = 180° $\quad$[Substituting the value of y]

⇒ x° = 180° - 75°

⇒ x° = 105°

x° = 105° and y° = 75°

(ii)

Given:

AB || CD

∠EAB = 130°

∠BCD = 70°

Since, EF is a straight line:

∠EAB + ∠BAC = 180° $\quad$[Linear pair]

⇒ 130° + ∠BAC = 180° $\quad$[Substituting the value of ∠EAB]

⇒ ∠BAC = 180° - 130°

⇒ ∠BAC = 50°

∠DCF = ∠BAC $\quad$[Corresponding angles]

⇒ x° = ∠BAC

⇒ x° = 50° $\quad$[Substituting the value of ∠BAC]

Since EF is a straight line:

∠BCA + ∠BCF = 180° $\quad$[Linear pair]

⇒ y° + (x° + 70°) = 180°

⇒ y° + 50° + 70° = 180° $\quad$[Substituting the value of x]

⇒ y° + 120° = 180°

⇒ y° = 180° - 120°

⇒ y° = 60°

In △ABC, we know that the sum of interior angles is 180°.

∴ ∠BAC + ∠ABC + ∠BCA = 180°

⇒ 50° + z° + y° = 180°

⇒ 50° + z° + 60° = 180° $\quad$[Substituting the value of y]

⇒ z° + 110° = 180°

⇒ z° = 180° - 110°

⇒ z° = 70°

x° = 50°, y° = 60° and z° = 70°

(iii)

Given:

∠CQP = 120°

∠DRP = 115°

Since, CD is a straight line:

∠CQP + ∠PQR = 180° $\quad$[Linear pair]

⇒ 120° + ∠PQR = 180° $\quad$[Substituting the value of ∠CQP]

⇒ ∠PQR = 180° - 120°

⇒ ∠PQR = 60°

AB || CD and PQ is a transversal:

∠APQ = ∠PQR $\quad$[Alternate angles]

⇒ x° = ∠PQR

⇒ x° = 60° $\quad$[Substituting the value of ∠PQR]

Since, CD is a straight line:

∠PRD + ∠PRQ = 180° $\quad$[Linear pair]

⇒ 115° + ∠PRQ = 180° $\quad$[Substituting the value of ∠PRD]

⇒ ∠PRQ = 180° - 115°

⇒ ∠PRQ = 65°

AB || CD and PR is a transversal:

∠BPR = ∠PRQ $\quad$[Alternate angles]

⇒ z° = ∠PRQ

⇒ z° = 65° $\quad$[Substituting the value of ∠PRQ]

In △PQR, we know that the sum of interior angles is 180°.

∴ ∠PQR + ∠QPR + ∠PRQ = 180°

⇒ 60° + y° + 65° = 180°

⇒ y° + 125° = 180°

⇒ y° = 180° - 125°

⇒ y° = 55°

x° = 60°, y° = 55° and z° = 65°

(iv)

Given:

∠DCE = 100°

∠BCA = 30°

Since, AE is a straight line:

∠BCA + ∠BCD + ∠DCE = 180°

⇒ 30° + z° + 100° = 180°

⇒ z° + 130° = 180°

⇒ z° = 180° - 130°

⇒ z° = 50°

AB || CD and BC is a transversal:

∠ABC = ∠BCD $\quad$[Alternate angles]

⇒ x° = z°

⇒ x° = 50°

In △ABC, we know that the sum of interior angles is 180°.

∴ ∠ABC + ∠BAC + ∠BCA = 180°

⇒ x° + y° + 30° = 180°

⇒ 50° + y° + 30° = 180°

⇒ y° + 80° = 180°

⇒ y° = 180° - 80°

⇒ y° = 100°

x° = 50°, y° = 100° and z° = 50°

(v)

From figure,

∠BAD = 35°

∠ECD = 75°

AB || CD and AD is a transversal:

∠ADC = ∠BAD $\quad$[Alternate angles]

∠ADC = 35°

∴ ∠CDE = 35° $\quad$[Both are same angles i.e., ∠D]

In △CDE, we know that the sum of interior angles is 180°.

∴ ∠ECD + ∠CDE + ∠DEC = 180°

⇒ 75° + 35° + x° = 180°

⇒ x° + 110° = 180°

⇒ x° = 180° - 110°

x° = 70°

(vi)

Through M, draw a line EMF such that EF || AB || CD.

This line splits the angle x into two parts:

Let ∠EMB = z° and ∠EMD = y°

Now, EF || CD and MD is a transversal.

∴ The sum of co-interior angles is 180°.

∴ ∠CDM + ∠DME = 180°

⇒ 130° + y° = 180°

⇒ y° = 180° - 130°

⇒ y° = 50°

Again, Now, EF || AB and MB is a transversal.

∴ The sum of co-interior angles is 180°.

∴ ∠ABM + ∠BME = 180°

⇒ 150° + z° = 180°

⇒ z° = 180° - 150°

⇒ z° = 30°

x° = y° + z°

⇒ x° = 50° + 30°

x° = 80°

(vii)

Through M, draw a line EMF such that EF || AB || CD.

This line splits the angle at M into two parts:

Let ∠EMB = y° and ∠EMD = (90 - y)°

Now, EF || CD and MD is a transversal.

∴ The sum of co-interior angles is 180°.

∴ ∠CDM + ∠DME = 180°

⇒ 140° + (90 - y)° = 180°

⇒ 140° + 90° - y° = 180°

⇒ 230° - y° = 180°

⇒ y° = 230° - 180°

⇒ y° = 50°

Again, EF || AB and MB is a transversal.

∴ The sum of co-interior angles is 180°.

∴ ∠ABM + ∠BME = 180°

⇒ x° + y° = 180°

⇒ x° + 50° = 180°

⇒ x° = 180° - 50°

x° = 130°

(viii)

Through M, draw a line EMF such that EF || AB || CD.

This line splits the angle at M into two parts:

Let ∠AME = y° and ∠CME = z°

Now, EF || AB and AM is a transversal.

∠AME = ∠BAM $\quad$[Alternate angles]

∴ y° = 40°

Again, EF || CD and CM is a transversal.

∠CME = ∠MCD $\quad$[Alternate angles]

∴ z° = 50°

x° = y° + z°

⇒ x° = 40° + 50°

x° = 90°

In each of the following figures AB || CD. Find the unknown angles, giving reasons.

(i)

(ii)

Answer

(i)

Given:

AB || CD

Since, AB is a straight line:

x° + 100° = 180° $\quad$[Linear pair]

⇒ x° = 180° - 100°

⇒ x° = 80°

y° = 120° $\quad$[Vertically opposite angles]

AB || CD and GH is a transversal:

t° = x° $\quad$[Alternate angles]

∴ t° = 80°

Let the angle opposite to z° be p°.

In quadrilateral ABCD, sum of interior angles is 360°.

∴ 120° + 100° + 80° + p° = 360°

⇒ 300° + p° = 360°

⇒ p° = 360° - 300°

⇒ p° = 60°

z° = p° $\quad$[Alternate angles]

∴ z° = 60°

x° = 80°, y° = 120°, z° = 60° and t° = 80°

(ii)

Given:

AB || CD

t° = 120° $\quad$[Vertically opposite angles]

AB || CD and EF is a transversal:

y° = t° $\quad$[Alternate angles]

∴ y° = 120°

x° = 50° $\quad$[Exterior alternate angles]

Since, AB is a straight line:

x° + z° = 180° $\quad$[Linear pair]

⇒ 50° + z° = 180°

⇒ z° = 180° - 50°

⇒ z° = 130°

x° = 50°, y° = 120°, z° = 130° and t° = 120°

In the given figure l || m and p || q. Find the angles x, y, z.

Answer

Consider p || q and m is a transversal:

x° = 70° $\quad$[Alternate angles]

Since, m is a straight line:

x° + z° = 180° $\quad$[Linear pair]

⇒ 70° + z° = 180°

⇒ z° = 180° - 70°

⇒ z° = 110°

Now, consider l || m and p is a transversal:

y° = x° $\quad$[Alternate angles]

∴ y° = 70°

x° = 70°, y° = 70° and z° = 110°

In the given figure, AB || CD || EF. Find x, y and z.

Answer

Consider AB || CD and AD is a transversal:

x° = 140° $\quad$[Alternate angles]

Since, x°, y° and 120° meet at same point D and form a complete angle:

x° + y° + 120° = 360°

⇒ 140° + y° + 120° = 360° $\quad$[Substituting the value of x]

⇒ 260° + y° = 360°

⇒ y° = 360° - 260°

⇒ y° = 100°

y° and z° are co-interior angles.

Co-interior angles are supplementary:

∴ y° + z° = 180°

⇒ 100° + z° = 180° $\quad$[Substituting the value of y]

⇒ z° = 180° - 100°

⇒ z° = 80°

x° = 140°, y° = 100° and z° = 80°

In the given figure, AB || CD || EF and AD || BE. Find x, y and z.

Answer

Consider AB || CD and AD is a transversal:

y° = 115° $\quad$[Alternate angles]

∠BAD and ∠ABE are co-interior angles.

Since, line GB intersects at point B, therefore ∠ABE = (x° + 30°)

Co-interior angles are supplementary:

∴ ∠BAD + ∠ABE = 180°

⇒ 115° + x° + 30° = 180° $\quad$[Substituting the value of ∠ABE]

⇒ x° + 145° = 180°

⇒ x° = 180° - 145°

⇒ x° = 35°

Now, consider AB || EF and BE is a transversal:

∠BEF = ∠ABE $\quad$[Alternate angles]

⇒ ∠BEF = x° + 30° $\quad$[Substituting the value of ∠ABE]

⇒ ∠BEF = 35° + 30° $\quad$[Substituting the value of x]

⇒ ∠BEF = 65°

∴ z° = 65°

x° = 35°, y° = 115° and z° = 65°

In the given figure, l || m || n and p || q || r. Find the angles x, y, z and t.

Answer

Consider, l || n:

x° and 100° are co-interior angles.

Co-interior angles are supplementary:

∴ x° + 100° = 180°

⇒ x° = 180° - 100°

⇒ x° = 80°

y° = x° $\quad$[Corresponding angles]

∴ y° = 80°

Consider p || r and n is a transversal:

z° = x° $\quad$[Alternate angles]

∴ z° = 80°

Now, consider l || m and r is a transversal:

t° = 100° $\quad$[Alternate angles]

x° = 80°, y° = 80°, z° = 80° and t° = 100°

In the given figure, l || m and p || q. Find the angles x, y, z and t.

Answer

The angles at the top intersection point (where lines l, q and the vertical line meet) must sum to 360° because they form a complete circle.

270° + 40° + x° = 360°

⇒ 310° + x° = 360°

⇒ x° = 360° - 310°

⇒ x° = 50°

Consider p || q:

z° = x° $\quad$[Corresponding angles]

∴ z° = 50°

Similarly,

y° = 40° $\quad$[Corresponding angles]

Since, p is a straight line:

z° + t° = 180° $\quad$[Linear pair]

⇒ 50° + t° = 180° $\quad$[Substituting the value of z]

⇒ t° = 180° - 50°

⇒ t° = 130°

x° = 50°, y° = 40°, z° = 50° and t° = 130°

In the given figure, AB || CD. Find the angles x, y and z.

Answer

From figure,

∠ECF = 90°, ∠CEF = 30°, ∠AFD = 80°

In △ECF, the sum of interior angles is 180°.

∴ ∠ECF + ∠CEF + ∠EFC = 180°

⇒ 90° + 30° + x° = 180°

⇒ 120° + x° = 180°

⇒ x° = 180° - 120°

⇒ x° = 60°

Consider, AB || CD:

z° and 80° are co-interior angles.

Co-interior angles are supplementary:

∴ z° + 80° = 180°

⇒ z° = 180° - 80°

⇒ z° = 100°

Since CD is a straight line, the sum of all angles on one side of a straight line at a point is 180°.

∴ x° + y° + 80° = 180°

⇒ 60° + y° + 80° = 180°

⇒ y° + 140° = 180°

⇒ y° = 180° - 140°

⇒ y° = 40°

x° = 60°, y° = 40° and z° = 100°

In the given figure, l || m and p || q. Find the angles x, y and z.

Answer

Consider p || q and l is a transversal:

x° = 100° $\quad$[Alternate angles]

Consider l || m:

z° = 100° $\quad$[Corresponding angles]

Let the angle vertically opposite to y° be t°.

Now, t° and z° are co-interior:

∴ t° + z° = 180°

⇒ t° + 100° = 180°

⇒ t° = 180° - 100°

⇒ t° = 80°

y° = t° $\quad$[Vertically opposite angles]

∴ y° = 80°

x° = 100°, y° = 80° and z° = 100°

In each of the following figures, two lines AB and CD are cut by a transversal EF. In each case, find whether AB || CD or not. Give reasons in support of your answer.

(i)

(ii)

(iii)

(iv)

(v)

Answer

(i)

In the given figure,

The angle vertically opposite to 40° is 40°.

Now, consider the pair of co-interior angles: 130° and 40°.

Sum = 130° + 40° = 170°

Since the sum of co-interior angles is not 180°.

AB and CD are not parallel.

(ii)

At the first intersection, the interior angle adjacent to 100° is 180° - 100° = 80° (Linear pair).

This 80° angle and the given 80° angle at the second intersection are corresponding angles.

Since 80° = 80°, the corresponding angles are equal.

AB and CD are parallel.

(iii)

The sum of angle adjacent to 120° and 120° is 180° because they form linear pair.

So,

Adjacent angle = 180° - 120° = 60°

This 60° angle and the given 60° angle are exterior alternate angles.

Since 60° = 60°, the external alternate angles are equal.

AB and CD are parallel.

(iv)

From the figure we have,

AD is a transversal

∠BAD = 50° and ∠ADC = 40°

These form a pair of interior alternate angles.

But 50° ≠ 40°

Since alternate angles are not equal,

AB and CD are not parallel.

(v)

The angles 75° and 100° are a pair of co-interior angles.

Co-interior angles are supplementary.

Sum = 75° + 100° = 175°

Since the sum of co-interior angles is not 180°,

AB and CD are not parallel.