Directions: Study the following questions carefully and choose the right answer.
Important for :
1
In the given figure AB || CD, ∠A = 128°, ∠E = 144°. Then, ∠FCD is equal to :

» Explain it
D
As per the given figure,

Through E draw EE’ || AB || CD.

Then, ∠AEE’ = 180° ‒ ∠BAE = (180° ‒ 128°) = 52°.

(Interior angles on the same side of the transversal are supplementary.)

Now, ∠E’EC = (144° ‒ 52°) = 92°.

∠FCD = ∠E’EC = 92° (Corr. ∠s).

Hence, option D is correct.

2
In the trapezium PQRS, QR || PS, ∠Q = 90°, PQ = QR and ∠PRS = 20°. If ∠TSR = θ, then the value of θ is :

» Explain it
C
In the given figure,

PQ = QR and ∠PQR = 90° ⇒ ∠QPR = ∠QRP = 45°.

∴ ∠QRS = (45° + 20°) = 65°.

∴ θ = ∠QRS = 65° (alt. ∠s)

Hence, option C is correct.

3
In the adjoining figure, ∠ABC = 100°, ∠EDC = 120° and AB || DE. Then, ∠BCD is equal to :

» Explain it
C
In the given figure,

Produce AB to meet CD at F.

∠BFD = ∠EDF = 120° (alt. ∠s)

∠BFC = (180° ‒ 120°) = 60°.

∠CBF = (180° ‒ 100°) = 80°.

∴ ∠BCF = 180° ‒ (60° + 80°) = 40°.

Hence, option C is correct.

4
In the given figure, AB || CD, ∠ABO = 40° and ∠CDO = 30°. If ∠DOB = x°, then the value of x is :

» Explain it
C
In the given figure,

Through O draw EOF parallel to AB & so to CD.

∴ ∠BOF = ∠ABO = 40° (alt. ∠s)

Similarly, ∠FOD = ∠CDO = 30° (alt. ∠s)

∴ ∠BOD = (40° + 30°) = 70°.

So, x = 70°.

Hence, option C is correct.

5
In the given figure, AB || CD, m∠ABF = 45° and m∠CFC = 110°. Then, m∠FDC is :

» Explain it
A
As in the given figure,

∠FCD = ∠FBA = 45° (alt. ∠s)

∴ ∠FDC = 180° ‒ (110° + 45°) = 25°.

Hence, option A is correct.