∠CAB + ∠ABC + ∠ACB = 180°

55° + 65° + ∠ACB = 180°

120° + ∠ACB = 180°

∠ACB = 60°

∠ACD + ∠ACB = 180°

x + 60° = 180°

x = 120°

So, the value of x is 120°.

In ΔABC, ∠CBD = ∠ACB + ∠CAB

m ∠CBD = 50 + 30

m ∠CBD = 80

z + 60 = 180

z = 180 - 60 = 120

The sum of the measures of the angles of a triangle is 180.

y + 60 + 90 = 180

y = 180 - (60 + 90) = 30

So, z + y = 120 + 30 = 150

∠ACB = 180^o - (55^o + 65^o) = 60^o

∠ACD = x^o = 180^o - 60^o = 120^o

So, the value of x is 120.

In ΔABC, m∠ABC = 54 and m∠BAC = 86

Since side BC is extended to the point D, exterior angle of ∠ACB is ∠ACD.

The measure of each exterior angle of a triangle equals the sum of the measures of its two remote interior angles.

m∠ABC + m∠BAC = m∠ACD

54 + 86 = m∠ACD

m∠ACD = 140

So, the measure of the marked exterior angle of the triangle is 140.

From the figure, ∠BAC = 90° and ∠ABD = 108°.

The measure of each exterior angle of a triangle equals the sum of the measures of its two remote interior angles.

∠BAC + ∠ACB = ∠ABD

90° + (x - 54)° = 108°

36° + x = 108°

⇒ x = 72°

So, the value of x in the figure is 72°.

From the figure, ∠CAB = 63° and ∠ABD = 136°.

The measure of each exterior angle of a triangle equals the sum of the measures of its two remote interior angles.

∠CAB + ∠ACB = ∠ABD

63° + y° = 136°

∠ACB = y° = 136° - 63°

So, the measure of y is 73°.

If one side of a triangle is produced, then the exterior angle so formed is equal to the sum of the interior opposite angles.

∠ACD = ∠BAC + ∠ABC

⇒ ∠ACD = 30° + x°

50° = 30° + x°

x° = 50° - 30°

⇒ x = 20°

Therefore, the measure of x is 20°

In ΔABC, ∠ABC = 52° and ∠BAC = 89°

Since side BC is extended to the point D, exterior angle of ∠ACB is ∠ACD.

The measure of each exterior angle of a triangle equals the sum of the measures of its two remote interior angles.

∠ABC + ∠BAC = ∠ACD

52° + 89° = ∠ACD

∠ACD = 141°

So, the measure of the marked exterior angle of the triangle is 141°.