Test: Lines And Angles- 2 - Class 9 MCQ

∠4 = ∠8 [ corresponding angles are equal]
∠7 + ∠8 = 180 [replacinig ∠8 by ∠4 ]
∠7 + ∠4 = 180

∵ In ∆ABC,
AB = AC
∴ ∠B = ∠C    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABC,
∠A + ∠B + ∠C = 180°
| Sum of all the angles of a triangle is 180°
⇒ 90° + ∠B + ∠C = 180°
| ∵ ∠A = 90° (given)
⇒    ∠B + ∠C = 90°    ...(2)
From (1) and (2), we get
∠B = ∠C = 45°.

Let the interior angles be x and 3x 

We know that exterior angle of triangle is equal to sum of interior opposite angles.

⇒ x+3x=80^∘

⇒ 4x=80^∘

⇒ x=20^∘

So the angles are 

x=20^∘

3x=3×20^∘

= 60∘

Step-by-Step Explanation:

1. Draw Parallel Line: Draw a line EF parallel to AB and CD through the point E.

2. Identify Angles:

   • ∠BXE = 120° is given.
   • ∠XED = 140° is also given.
   • Let ∠EFX be the angle formed by the line EF and the transversal EX.

3. Corresponding Angles: Since EF || AB and AB || CD, the angles formed by the transversal EX with these parallel lines are corresponding angles. Therefore:

   • ∠BXE = ∠FEX = 120° (corresponding angles).
   • ∠XED = ∠EFX = 140° (corresponding angles).

4. Sum of Angles Around Point E: The angles around point E add up to 360°. We have:
   ∠FEX + ∠XED + ∠EFX + x = 360°Substituting the values:
   120° + 140° + x = 360°

5. Calculate x: Solve for x:
   260° + x = 360°
   x = 360° − 260°
   x = 100°
   Thus, by introducing the parallel line EF and using the properties of corresponding angles, we conclude that the value of x is 100°.

A+b =180(linear pair)

2x+3x=180

5x=180

x=180/5

x=36

a=2x=2*36=72

b=3x=3*36=108

b=d (vertical opposite angles are equal)

d=f (alternative interior angles are equal)

f=h (vertically opposite angles are equal)

So, h= 108

∠POR + ∠QOR = 180° (Linear pair axiom)
But:
∠POR = 4x
∠QOR = 2x
Therefore:
∴ 4x + 2x = 180°
6x = 180°
x = 30°
Thus, the value of x = 30°.

Given that a > b by one third of a right angle:
a = b + (1/3) × 90° → a = b + 30°
But a + b = 180° (Linear pair axiom):
b + 30° + b = 180°
2b + 30° = 180°
2b = 180° - 30°
2b = 150°
Thus:
b = 75°
a = 180° - 75° = 105°
Therefore, the measure of a is 105° and b is 75°.

∠AOC + ∠BOD = 70° (Given) ................... (i)
∠AOC + ∠COD + ∠BOD = 180° (Linear pair axiom)
∴ ∠COD = 180° - (∠AOC + ∠BOD)
= 180° - 70° [From (i)]
= 110°

∴ The measure of ∠COD = 110°.

Let the two interior angles on the same side of the transversal be 5x and 4x. The sum of these angles is 180°, as they are co-interior angles.

5x + 4x = 180°
9x = 180°
x = 180° / 9 = 20°

The smaller angle is 4x = 4 × 20 = 80°.

Given, AB ║ CD and AB ║ EF

so CD || EF

which means ∠ECD + ∠CEF = 180⁰ (corresponding angles)

∠ECD = 180 - 150 = 30⁰

since AB || CD so 

∠ABC= ∠BCD (alternate interior angles)

∠ABC = 30 + ∠ECD = 30 + 30 = 60⁰

The sum of all angles around a main point equals 360 degree , as it represents a full rotation around the point.

Final Answer: A: 360°.

The supplement of an angle is calculated as:
Supplement = 180° − Given angle

For 105°:
Supplement = 180° − 105° = 75°

Let the angle be x. The complement of x is 90° - x, and the supplement of x is 180° - x.

From the condition, "Six times its complement is 12° less than twice its supplement," we have:

6(90 - x) = 2(180 - x) - 12

Expanding both sides:
540 - 6x = 360 - 2x - 12

Simplify:
540 - 6x = 348 - 2x

Rearranging terms:
540 - 348 = 6x - 2x
192 = 4x

Solving for x:
x = 192 / 4 = 48°

The angle is 48°.