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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(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°