Lines & Angles - 1 - Free MCQ Practice Test with solutions, Class 10

Given that, BC || AE
∠CBA + ∠EAB = 180°
⇒ ∠EAB = 180° – 65° = 115°
∵   BC = AC
Hence, ΔABC is an isosceles triangle. 

⇒ ∠CBA = ∠CAB = 65°
Now,  ∠EAB = ∠EAC + ∠CAB
⇒ 115° = x + 65° ⇒ x = 50°.
Hence, option D is correct.

Given that,

∠ABC = 65° and ∠CDE = 15°

Here, ∠ABC + ∠TCB = 180°

(∵   AB || CD)

∠TCB = 180° – ∠ABC

∴ ∠TCB = 180° – 65° = 115°

∵ ∠TCB + ∠DCB = 180°

(Linear pair)

∴ ∠DCB = 65°

Now, in ΔCDE

∠CED = 180° – (∠ECD + ∠EDC)

(∵ ∠ECD = ∠BCD)

= 180° – (– 65° + 15° ) = 100°

∵ ∠DEC + ∠FEC = 180°

⇒ ∠FEC = 180° – 100° = 80°

Given that, AB = AE.

i.e. ΔABE an isosceles triangle.

∴ ∠ABE = ∠AEB = 65°

∵ ∠AEB + ∠AEF + ∠FEC = 180°

(straight line)

⇒ 65° + x° + 80° = 180°

∴ x° = 180° – 145° = 35°.

Hence, option B is correct.

∠PCT + ∠PCB = π
(Linear pair)

∠PCB = π – (π – b°) = b° ..... (i)

In ΔBPC,

∠PCB + ∠BPC + ∠PBC = π

∠PBC = π – ∠PCB – ∠BPC  =  π – b° – a° ..... (ii)

∵ ∠ABE + ∠EBC = π
(∵ ∠PBC = ∠EBC)      (linear pair)
∠ABE = π – ∠PBC = π – (π – b° – a°)  =  a° + b° ...(iii)

Now, in ΔABE

Sum of two interior angles = Exterior angle

∠EAB + ∠ABE = ∠BES ⇒ c° + b° + a° = x°

∴   x° = a° + b° + c°.

Hence, option C is correct.

Statements I and II are both true, because the locus of points which are equidistant from two parallel lines is a line parallel to both of them and draw mid way between them.

Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

Hence, option C is correct.

= 12 × Its circumference
= 12 × 2πr
∴  In 1 s distance travelled by the wheel =
(12 × 2πr) / 60 = (2 / 5) πr
∴ Angle = (Arc / Radius)
= (2 / 5πr) / r = 2π / 5
Which is the required angle.
Hence, option B is correct.

If one arm of an angle is respectively parallel to the arm of another angle formed with a common line between them, then the two angles are not equal but supplementary.

If l₁ || l₂ ⇒ ∠1 + ∠2 = 180°
(Supplementary)

Hence, option B is correct.

Given that,
(1 / 2) ∠A + (1 / 3) ∠C + (1 / 2) ∠B = 80°
⇒   3∠A + 2∠C + 3∠B = 480°
⇒   3(∠A + ∠B) + 2∠C = 480° ....(i)
Also, in ΔABC,
∠A + ∠B + ∠C = 180°
On multiplying both sides L.H.S. & R.H.S. by 3, we get
3(∠A + ∠B) + 3∠C = 540°  ..... (ii)
On subtracting Eq. (i) from Eq. (ii), we get
∠C = 60°.
Hence, option C is correct.

Complementary angles: Complementary angles are angle pairs whose measures sum to one right angle (90°).

So, the required angle will be 10°
180° = π radian
1º = (π / 180)
∴ 10º = {(π × 10) / 180} = (π / 18)
Hence, option C is correct.

∠ALC = ∠LCD = 60°
[∵  Alternate angles]
EC is the bisector of ∠LCD
∴ ∠ ECD = (1 / 2) × ∠ LCD
= (1 / 2) × 60º = 30º
∠CEF + ∠ECD = 180°
[∵  Pair of interior angles]
∠CEF + 30° = 180°
∠CEF = 180° – 30° = 150°
Hence, option C is correct.

As AP is a straight line and rays GO and GF stands on it.
∴ ∠AGO + ∠OGF + ∠PGF = 180°

⇒ (∠AGO + ∠PGF) + ∠OGF = 180°

⇒ 70° + ∠OGF = 180°

⇒ ∠OGF = 180° – 70° 

⇒ ∠OGF = 110°

As, OQ is a straight line, rays GF and GP stands on it.

∠OGF + ∠PGF + ∠PGQ = 180°

Putting the value of ∠OGF & ∠PGQ 

110° + ∠PGF + 40° = 180°

∠PGF = 180° – 150° = 30°
Hence, option C is correct.