Notes on Angles | Grade 7 > Compulsory Maths > Geometry:Angles | KULLABS.COM

Notes, Exercises, Videos, Tests and Things to Remember on Angles

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#### Angles

The amount of turn between two straight lines that have a common endpoint that is a vertex.

Different Pairs of angles

$$\angle$$AOB and$$\angle$$BOC have common vertex O and a common arm OB. They are called adjacent angles.

2. Linear Pair
$$\angle$$AOB and$$\angle$$BOC are a pair of adjacent angles. Their sum is a straight angle (180°).
i.e$$\angle$$AOB +$$\angle$$BOC = 180°
$$\angle$$AOB and$$\angle$$BOC are called a linear pair.

3. Vertically Opposite Angles
$$\angle$$AOC and$$\angle$$BOD are formed by the intersected line segments and they lie to the opposite side of the common vertex. They are called vertically opposite angles.$$\angle$$AOD and$$\angle$$BOC are another pairs of vertically opposite angles. Vertically opposite angles are always equal.
$$\therefore$$$$\angle$$AOC =$$\angle$$BOD and$$\angle$$AOD =$$\angle$$BOC.

4. Complementary Angles
The sum of$$\angle$$AOB and$$\angle$$BOC is a right angle (90°). i.e$$\angle$$AOB +$$\angle$$BOC = 90°
$$\angle$$AOB and$$\angle$$BOC are called complementary angles.
Here, complement of$$\angle$$AOB = 90° -$$\angle$$BOC
Complement of$$\angle$$BOC = 90° - AOB

5. Supplementary Angles
The sum of $$\angle$$AOB and$$\angle$$BOC is two right angles (180°).
i.e$$\angle$$AOB +$$\angle$$BOC = 180°
$$\angle$$AOB and$$\angle$$BOC are called supplementary angles.
Here, supplement of$$\angle$$AOB = 180° -$$\angle$$BOC
Supplement of$$\angle$$BOC = 180° -$$\angle$$AOB

Pairs of angles made by a transversal with lines

In the given figure, AB and CD and two parallel lines (AB//CD). PQ is the transversal that intersects AB at R and CD at S.

1. Exterior and alternate exterior angles
$$\angle$$a,$$\angle$$b,$$\angle$$c and$$\angle$$d are lying outside the parallel lines. They are called exterior angles.$$\angle$$a and$$\angle$$d are lying to the opposite side of the transversal. They are called alternate exterior angles.
The alternate exterior angles made by a transversal with parallel lines are always equal.
$$\therefore$$$$\angle$$a =$$\angle$$d and$$\angle$$b =$$\angle$$c
2. Interior and Co-interior angles
$$\angle$$a,$$\angle$$b,$$\angle$$c and$$\angle$$d are lying inside the parallel lines. They are called interior angles. $$\angle$$a and$$\angle$$c are the pair of interior angles lying to the same side of the transversal. They are called Co-interior angles. The sum of a pair of co-interior angles made by a transversal with parallel lines is always 180°.
$$\therefore$$$$\angle$$a + $$\angle$$c = 180° and$$\angle$$b +$$\angle$$d = 180°
3. Alternate angles
$$\angle$$a and$$\angle$$d are a pair of interior angles lying to the opposite side of a transversal and they are not adjacent to each other. They are called alternate angles.$$\angle$$b and$$\angle$$c are another pairs of alternate angles.
A pair of alternate angles made by a transversal with parallel lines is always equal.
$$\therefore$$$$\angle$$a =$$\angle$$d and$$\angle$$b =$$\angle$$c

4. Corresponding angles
$$\angle$$a is an exterior and$$\angle$$d is an interior angle lying to the same side of the transversal and they are not adjacent to each other.; They are called corresponding angles. b and d are another pairs of corresponding angles.
A pair of corresponding angles made by a transversal with parallel lines is always equal.

• Definition of angles
• Different pairs of angle
• Pairs of an angle made by transversal lines.
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#### Click on the questions below to reveal the answers

Solution:

Here, x° + (x+10)° = 90° [The sum of a pair of complementary angles]

or, 2x° = 90° - 10°

or, x° = $$\frac{80°}{2}$$ = 40°

$$\therefore$$ x° = 40° and (x+10)° = 40° + 10° = 50°

Solution:

Let the required supplementary angles be 3x° and 2x°.

$$\therefore$$ 3x° + 2x° = 180° [The sum of a pair of supplementary angles]

or, 5x° = 180°

x° = $$\frac{180°}{5}$$ = 36°

$$\therefore$$ 3x° = 3 × 36° = 108°

2x° = 2 × 36° = 72°

Solution:

Here, x° + 2x° + 3x° = 180° [Being the sum a straight angle]

or, 6x° = 180°

or, x° = $$\frac{180°}{6}$$ = 30°

$$\therefore$$ x° = 30°, 2x° = 2×30° = 60° and 3x° = 3×30° = 90°

Again, a° = x° = 30°, b = 2x° = 60° and c° = 3x° = 90° [Each pair is vertically opposite angles]

Solution:

w = 110° [Being vertically opposite angles]

x = w = 110° [Being alternate angles]

y = x = 110° [Being vertically opposite angles]

y + z = 180° [Being the sum of a pair of co-interior angles]

or, 110 + z = 180°

or, z = 180° - 110° = 70°

So, w = x= y = 110° and z = 70°

0%

68°
67°
69°
66°

73° and 107°
72° and 108°
75° and 106°
74° and 109°

70° and 80°
80° and 100°
70° and 90°
80° and 90°

30° and 150°
30° and 180°
20° and 100°
20° and 120°

vertex
base
angles
edges

90°
360°
240°
180°

90°
100°
95°
105°

240°
180°
360°
90°

26°
28°
24°
22°

240°
180°
90°
360°

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