Test: Lines & Angles - 2 - Grade 10 MCQ

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.

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.

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.

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.

As in the given figure,
∠FCD = ∠FBA = 45° (alt. ∠s)
∴ ∠FDC = 180° ‒ (110° + 45°) = 25°.
Hence, option A is correct.

As in the given figure,
∠ADB = 180° ‒ (110° + 30°) = 40°.
So, ∠BDC = (75° ‒ 40°) = 35°.

∴ ∠DBC = 180° ‒ (60° + 35°) = 85°.

∴ ∠BCE = ∠DBC = 85° (alt. ∠s).

So, x = 85°.
Hence, option C is correct.

Let the two angles are x and y. Therefore, as per the given information,
x ‒ y = 44° and 
x + y = 180°
[ As the total of supplementary angles is 180°]
On solving these two linear equations we get,
2x = 224, 
x = 112°. 
Therefore the other angle y = 180° ‒ 112° = 68°
Hence, option D is correct.

Here, AB and CD are two lines.

If two straight lines intersect, then opposite vertically angles are equal.
Hence, option B is correct.

From the given figure,
∠LOQ + ∠QOP + ∠POM = 180°
(straight line)
∴   (x° + 20°) + 50° + (x° – 10°) = 180°
⇒   2x° + 60° = 180° ⇒ 2x° = 120°
∴   x° = 60°
Hence, optuion B is correct.

Given that, EC || AB
∴    ∠ECO + ∠AOC = 180°

⇒   ∠AOC = 180° – 70° = 110°

∴    ∠BOD = ∠AOC = 110°

(alternate angle)

Now, in ΔOBD

∠BOD + ∠ODB + ∠DBO = 180°

∴   110° + 20° + x° = 180° ⇒ x° = 50°.

Hence, option D is correct.