r = 80^o (Corresponding angles)

CORRESPONDING ANGLES

r = 80^o (Corresponding angles)

CORRESPONDING ANGLES

r + 5k = 180 (Sum of angles on a straight line is 180)
But r = 80
80 + 5k = 180
5k = 180 - 80
5k = 100
Divide both sides by 5 so that k stands alone.

k = $\frac{100}{5}$ = 20

You have to use the property of alternate angles of parallel lines

Alternate angles are equal:

Sometimes called Z angles

We can often spot interior alternate angles by drawing a Z shape

From the diagram above, angle ABC = angle CDE = x (Alternate angles)

Angle BAC = y = DEC (Alternate angles)

Angle ABC = angle CDE = x (Alternate angles)

Angle ABC = 81^o, ∴ x = 81^o

You have to use the property of alternate angles of parallel lines

Alternate angles are equal:

Sometimes called Z angles

We can often spot interior alternate angles by drawing a Z shape

From the diagram above, angle ABC = angle CDE = x (Alternate angles)

Angle BAC = y = DEC (Alternate angles)

Angle ABC = angle CDE = x (Alternate angles)

Angle ABC = 81^o, ∴ x = 81^o

Since we know the angles DCE and CDE, we can calculate the angle DEC using the concept that angles in a triangle add up to 180^o

x = 81^o and DCE = 20^o

Sum of angles in triangle add up to 180^o

81 + 20 + angle DEC = 180

101 + DEC = 180

DEC = 180 - 101

DEC = 79

Since BAC = y = DEC, y = 79^o

Properties

Alternate Angles

Alternate angles are equal

From the diagram above, angle DCB and angle OBA are alternate angles

Hence angle DCB = OBA

Sum of angles in triangle

Sum of angles in a triangle add up to 180^o

A^o + B^o + C^o = 180^o

From the diagram in the question, angle x^o + angle OBA + angle BAO = 180^o (Sum of angles in triangle)

x^o + OBA^o + 20^o = 180^o

x^o = 180^o - 20^o - OBA^o

x^o = 160^o - OBA^o

Sum of angles on a straight line

x + y = 180 (sum of angles on a straight line add up to 180)

From the diagram in the question, 120 + angle DCB = 180 (Sum of angles on a straight line)

120 + DCB = 180

DCB = 180 - 120

DCB = 60

Angle DCB = OBA, alternate angles

If angle DCB = 60, then angle OBA is also 60

From x^o = 160^o - OBA^o

OBA = 60

x = 160 - 60

x = 100

x = 8

(9x + 8)^o = (9 x 8 + 8)^o

(9x + 8)^o = (72 + 8)^o

(9x + 8)^o = (80)^o

80^o is less than 90^o, hence the angle is acute

Concepts

A triangle which has two of its sides equal is called an isosceles triangle

Properties of Isosceles Triangle

1. The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.

2. The side opposite the vertex angle is called the base and base angles are equal.

3. The perpendicular from the vertex angle bisects the base and the vertex angle.

Applying property 2 to redraw the triangle in the question gives us the diagram below:

From the above properties, you will realize that angle CBA = BAC = 56^o

Applying the concept that says that angles on a straight line add up to 180^o

⇒ x+56 = 180

Make x the subject

⇒ x = 180 - 56

x = 124

Concepts

A triangle which has two of its sides equal is called an isosceles triangle

Properties of Isosceles Triangle

1. The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.

2. The side opposite the vertex angle is called the base and base angles are equal.

3. The perpendicular from the vertex angle bisects the base and the vertex angle.

Applying property 2 to redraw the triangle in the question gives us the diagram below:

From the above properties, you will realize that angle CBA = BAC = 56^o

Considering the basic properties of angles in a triangle below:

Applying the concept which says that the sum of all the angles in a triangle should be 180^o

⇒ 56+y+56 = 180

y+56+56 = 180

y+112 = 180

⇒ y = 180 - 112

y = 68^o