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