OR = OQ = Radius

∆ RQO is an isosceles triangle.

Isosceles triangle has the base angles the same.

∠ ORQ = ∠ OQR = 42°

∠ ROQ + 42° + 42° = 180° (sum of angles of a triangle)

∠ ROQ + 84 = 180

∠ ROQ = 180 - 84

∠ ROQ = 96°

QPR = $\frac{1}{2}$ x ∠ ROQ

QPR = $\frac{1}{2}$ x 96° = 48°

Draw an imaginary parallel line to cut t.

t = ∠ CGE + ∠ EGJ

∠ ACD = ∠ BCG = 35° (Vertically opposite angles)

∠ CGE = ∠ BCG = 35° (Alternate angles)

∠ HJG + ∠ IJG = 180° (Angles on a straight line)

∠ HJG = 140°

140° + ∠ IJG = 180°

∠ IJG = 180° - 140°

∠ IJG = 40°

∠ EGJ = IJG = 40° (Alternate angles)

t = ∠ CGE + ∠ EGJ

t = 35° + 40°

t = 75°

|OS| = |OQ| = Radius of the circle

∆ QOS is an isosceles triangle hence the base angles are the same.

∠ OSQ = ∠ OQS

∠ QOS + ∠ OSQ + ∠ OQS = 180°(Sum of angles of a triangle)

108° + ∠ OSQ + ∠ OQS = 180°

∠ OSQ + ∠ OQS = 180° - 108°

∠ OSQ + ∠ OQS = 72°

∠ OSQ = ∠ OQS = $\frac{\mathrm{72°}}{2}$ = 36°

From alternate segment theorem,

Draw an imaginary angle subtended by the arc QS

From the alternate segment theorem, ∠ QTS = ∠ SQR

∠ QTS = $\frac{\mathrm{\angle \; QOS}}{2}$ = $\frac{\mathrm{108°}}{2}$ = 54°

∠ QTS = ∠ SQR = 54°

From alternate segment theorem,

∠ MNP = ∠ QMP = 73°

∠ MNP + ∠ MPN + ∠ PMN = 180°(Sum of angles in a triangle)

73° + ∠ MPN + 58° = 180°

∠ MPN + 131° = 180°

∠ MPN = 180° - 131°

∠ MPN = 49°

PS = RS

∆ PSR is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ SPR = ∠ SRP

∠ PSR = 90°

∠ SPR + ∠ SRP + ∠ PSR = 180° (Sum of angles in a triangle)

∠ SPR + ∠ SRP + 90° = 180°

∠ SPR + ∠ SRP = 180° - 90°

∠ SPR + ∠ SRP = 90°

∠ SPR = ∠ SRP = $\frac{\mathrm{90°}}{2}$ = 45°

∠ PQS = ∠ SRP = 45°(Angles subtended by the same arc PS)

Sum of angles on a straight line is 180^o

p + ∠ BAC = 180^o

∠ BAC = 180^o - p

q + ∠ BCA = 180^o

∠ BCA = 180^o - q

Sum of interior angles of a triangle is 180^o

s + ∠ BAC + ∠ BCA = 180^o

s + (180^o - p) + (180^o - q) = 180^o

s + 180^o - p + 180^o - q = 180^o

s + 180^o + 180^o - p - q = 180^o

s + 360^o - p - q = 180^o

s = 180^o - 360^o + p + q

s = -180^o + p + q

s = p + q - 180^o

But p + q = 250^o

s = 250^o - 180^o

s = 70^o

Inscribed angles subtended by the same arc are equal.

∠ PMQ = ∠ QNP = 57^o

∠ MPN = ∠ MQN = 43^o

Sum of interior angles of a triangle is 180^o

57^o + ∠ MPN + ∠ PON = 180^o

∠ MPN = 43^o

57^o + 43^o + ∠ PON = 180^o

100^o + ∠ PON = 180^o

∠ PON = 180^o - 100^o

∠ PON = 80^o

Sum of angles on a straight line is 180^o

y + ∠ PON = 180^o

But ∠ PON = 80^o

y + 80^o = 180^o

y = 180^o - 80^o

y = 100^o

Alternate Angles

Project the parallel lines to see which angles are alternate anges.

x = a + b

a = 35^o (Alternate angles)

b = c (Alternate angles)

130^o + c = 180^o (Sum of angles on a straight line)

c = 180^o - 130^o

c = 50^o

b = 50^o

x = a + b

x = 35^o + 50^o

x = 85^o

Angles subtended by the diameter (or semi-circle) is 90°.

∠MNP = 90°

Sum of angles on a straight line is 180°

(5x + 18)° + ∠NPM = 180°

5x + 18 + ∠NPM = 180

∠NPM = 180 - 5x - 18

∠NPM = 162 - 5x

∠NPM + ∠MNP + ∠NMP = 180° (sum of angles in a triangle)

162 - 5x + 90 + 2x = 180

162 + 90 - 5x + 2x = 180

252 - 3x = 180

252 - 180 = 3x

72 = 3x

Divide both sides by 3

x = $\frac{72}{3}$ = 24°

Angles subtended by the diameter (or semi-circle) is 90°.

∠YXW = 90°

26° + ∠ZXW = ∠YXW = 90°

26° + ∠ZXW = 90°

∠ZXW = 90° - 26°

∠ZXW = 64°

Base angles of isosceles triangle are the same.

∠WZX = ∠XWZ

∠WZX + ∠XWZ + ∠ZXW = 180°(Sum of interior angles of a triangle)

∠WZX + ∠XWZ + 64° = 180°

∠WZX + ∠XWZ = 180° - 64°

∠WZX + ∠XWZ = 116°

∠WZX = ∠XWZ = $\frac{\mathrm{116°}}{2}$ = 58°

Sum of two opposite interior angles of a triangle is equal to it's angle exterior.

∠XYZ + ∠YXZ = ∠WZX

∠XYZ + 26° = 58°

∠XYZ = 58° - 26°

∠XYZ = 32°

Inscribed angles subtended by the same are equal.

∠PSQ = ∠PRQ = 61^o

Sum of angles in are triangle is 180^o

∠QPS + ∠PSQ + ∠PQS = 180^o

∠QPS + 61 + 32 = 180

∠QPS + 93 = 180

∠QPS = 180 - 93

∠QPS = 87^o

Vertically opposite angles are the same.

Sum of angles in a triangle is 180^o

3y + 2y + 3y + 12 = 180
8y = 180 - 12
8y = 168

Divide both sides by 8

y = $\frac{168}{8}$

y = 21^o

The angle adjacent to 110° is (180 - 110)° (Angles on a straight line sum up to 180°)

(x + 4y)° = The angle adjacent to 110° (Alternate angles are the same)

(x + 4y)° = (180 - 110)°
x + 4y = 70 ------ (equation 1)

(x + 4y)° = (2x + y)°(Vertically opposite angles are the same).

Since x + 4y = 70, 2x + y is also 70

2x + y = 70 ------ (equation 2)

Solve equation 1 and 2 simultaneously.

Get rid of x by multiplying equation 1 by 2 and subtracting equation 2 from equation 3.

x x 2 + 4y x 2 = 70 x 2
2x + 8y = 140 ------ (equation 3)

Equation 3 - Equation 2

7y = 70

Divide both sides by 7

y = $\frac{70}{7}$

y = 10

Subtitute y = 10 into equation 1

x + 4 x 10 = 70
x + 40 = 70
x = 70 - 40
x = 30

x + y = 30 + 10
x + y = 40

Angles subtended by the diameter (or semi-circle) is 90^o.

∠PQR = 90^o

∠PQS + ∠RQS = ∠PQR = 90^o

∠RQS = 23^o

∠PQS + 23^o = 90^o

∠PQS = 90^o - 23^o

∠PQS = 67^o

Since triangle PQS is isosceles triangle, the base angles are the same.

∠PQS = ∠PSQ = 67^o

Sum of angles in a triangle is 180^o

x + ∠PQS + ∠PSQ = 180^o

x + 67^o + 67^o = 180^o

x + 134^o = 180^o

x = 180^o - 134^o

x = 46^o

Central angle is twice any inscribed angle subtended by the same arc.

∠ EPG = 2 x ∠ EFG

∠ EPG = 2 x 142^o

∠ EPG = 284^o

∠ EPG + x = 360^o (Sum of angles of a circle)

284^o + x = 360^o

x = 360^o - 284^o

x = 76^o

Sum of interior angles of a quadrilateral is 360^o

Note: sum of interior angles = (n - 2) x 180^o, where n is the number of sides.

63^o + 142^o + ∠ FEP + x = 360^o

63^o + 142^o + ∠ FEP + 76^o = 360^o

∠ FEP + 281^o = 360^o

∠ FEP = 360^o - 281^o

∠ FEP = 79^o

Angles subtended by the diameter (or semi-circle) is 90^o.

∠PRQ = 90^o

∠PRQ = ∠QRS + ∠PRS

∠PRS = ∠PRQ - ∠QRS

∠PRS = 90^o - ∠QRS

36^o + 64^o + ∠QRS = 180^o (sum of angles in a triangle)

100^o + ∠QRS = 180^o

∠QRS = 180^o - 100^o

∠QRS = 80^o

∠PRS = 90^o - 80^o

∠PRS = 10^o

Right angle means 90°

Sum of angles of a regular polygon = (n - 2) x 180°

Number of sides (n) of the polygon = 4

Sum of angles of a regular polygon = (4 - 2) x 180°

Sum of angles of a regular polygon = 2 x 180°

Sum of angles of a regular polygon = 360°

∠ADC = ∠EDF = 70°

90° + 90° + t + 70° = 360°

t + 250° = 360°

t = 360° - 250°

t = 110°

Isosceles triangles (triangle with two sides equal) have the base angles the same.

∠ YXZ = ∠ XZY

Let ∠ YXZ = ∠ XZY = x

x + x + 40° = 180° (Sum of angles in a triangle)

2x = 180° - 40°

2x = 140°

Divide both sides by 2.

x = $\frac{\mathrm{140°}}{2}$ = 70°

∠ YXZ = ∠ XZY = 70°

∠ XZY + ∠XZT = 180° (Sum of angles on a straight line)

70° + ∠XZT = 180°

∠XZT = 180° - 70°

∠XZT = 110°

Note:

1.

Central angle is twice any inscribed angle subtended by the same arc.

∠ POR = 2 x ∠PQR = 2 x 72° = 144°

2.

∠ POT = ∠ OPS (Alternate angles).

∠ POT + ∠ POR = 180° (Angles on a straight line)

∠ POT + 144° = 180°

∠ POT = 180° - 144°

∠ POT = 36°

∠ POT = ∠ OPS = 36°

b + 110 = 180 (Sum of angles on a straight line)

b = 180 - 110

b = 70°

a = b = 70° (Alternate angles)

a = y = 70° (Vertically opposite angles)

c = 35 (Vertically opposite angles)

c = d = 35 (Alternate angles)

x + d = 180 (Sum of angles on a straight line)

x + 35 = 180

x = 180 - 35

x = 145°

x + y = 70° + 145° = 215°

Apply the alternate segment theorem.

∠ QRN = 64°

The base angles of isosceles triangles (triangle with two sides equal) are the same.

∠ RQN = ∠ RNQ

Let ∠ RQN = ∠ RNQ = x

x + x + 64 = 180 (Sum of angles of a triangle)

2x = 180 - 64

2x = 116

Divide both sides by 2.

x = $\frac{116}{2}$ = 58°

∠ RQN = ∠ RNQ = 58°

From the the alternate segment theorem, ∠ RQN = t

∠ RQN = t = 58°

In cyclic quadrilateral, the opposite angles add up to 180°

x + (x + y) = 180

x + x + y = 180

2x + y = 180 ----- eqn 1

(2y - 30) + x = 180

2y - 30 + x = 180

x + 2y = 180 + 30

x + 2y = 210 ----- eqn 2

Get rid of y.

eqn 1 x 2

4x + 2y = 360 ----- eqn 3

eqn 3 - eqn 2

3x = 150

Divide both sides by 3

x = $\frac{150}{3}$ = 50°

1. A diameter subtends an angle of 90° at the circumference of the circle

In the diagram EC is a diameter

∠ EBC = 90° (subtended by the diameter)

∠ ABE + ∠ EBC = ∠ ABC

∠ ABE + 90° = 158°

∠ ABE = 158° - 90°

∠ ABE = 68°

2. Angles in the same segment are equal

∠ ABE = ∠ ADE = 68°(Angles on the same segment of chord AE)

Draw an imaginary third parallel line.

∠ QPO = ∠ TOP = x° (Alternate angles)

∠ RSO = ∠ TOS = 68° (Alternate angles)

x° + 68° + 246° = 360° (Sum of angles at centre)

x° + 314° = 360°

x° = 360° - 314°

x° = 46°

x = 46

The midpoint divides XY into two equal lengths, 3 cm each.

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan ∠ XZT = $\frac{\mathrm{XT}}{\mathrm{ZT}}$

XT = 3 cm but ZT is unknown.

You can use Pythagoream theorem to calculate ZT.

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypotenuse (c)

ZT² + XT² = XT²

ZT² + 3² = 6²

ZT² + 9 = 36

ZT² = 36 - 9

ZT² = 27

Take square root of both sides.

ZT = $\sqrt{\mathrm{27}}$

27 = 9 x 3

ZT = $\sqrt{\mathrm{9\; x\; 3}}$

But $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

ZT = $\sqrt{9}$ x $\sqrt{3}$

$\sqrt{9}$ = 3

ZT = 3 x $\sqrt{3}$ = 3$\sqrt{3}$

tan ∠ XZT = $\frac{3\sqrt{3}}{3}$

3 cancel out 3.

tan ∠ XZT = $\sqrt{3}$

∠ MNP = ∠ NPQ = 2x = (3x - 50°) (Alternate angles)

2x = (3x - 50°)

2x = 3x - 50°

50° = 3x - 2x

50° = x

x = 50°

∠ NPQ = 2x = 2 x 50° = 100°

Alternatively,

∠ NPQ = 3x - 50°

∠ NPQ = 3 x 50° - 50°

∠ NPQ = 150° - 50°

∠ NPQ = 100°

OP = OQ (Both are radius)

Since two sides of ∆OPQ are the same (isosceles), the base angles are the same.

∠ OPQ = ∠ OQP = 48°

Sum of two opposite interior angles of a triangle is equal to it's angle exterior.

m = ∠ OPQ + ∠ OQP

m = 48° + 48° = 96°

Notes:

Angles subtended by the diameter (or semi-circle) is 90°.

∠ QPS = 90°

Angles subtended by the same arc are the same.

From arc SP,

∠ PRS = ∠ PQS = 37° (Angles subtended by the same arc)

∠ PQS + ∠ QPS + ∠ PSQ = 180° (Sum of angles in a triangle)

37° + 90° + ∠ PSQ = 180°

127° + ∠ PSQ = 180°

∠ PSQ = 180° - 127°

∠ PSQ = 53°

m = $\frac{1}{2}$(x + y + z) ------- eqn 1

x + y + m + z = 180° (Sum of angles on a straight line is 180°)

x + y + z = 180° - m

Substitute (x + y + z) = 180° - m into ----- eqn 1

m = $\frac{1}{2}$(180° - m)

Get rid of the fraction by multiplying both sides by the denominator (2).

2 x m = $\frac{1}{2}$(180° - m) x 2

2m = 180° - m

2m + m = 180°

3m = 180°

Divide both sides by 3

m = $\frac{\mathrm{180°}}{3}$ = 60°

Method I

Drawing a third parallel line.

∠ QUT = ∠ UTO = m° (Alternate angles)

∠ SVT = ∠ VTO = n° (Alternate angles)

∠ UTV = ∠ UTO + ∠ VTO = m° + n°

Angles subtended by the diameter (or semi-circle) is 90°.

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

∠ UTV = m° + n° = 90°

Method II

Apply the alternate segment theorem.

∠ QUT = ∠ UVT = m° (alternate segment)

∠ SVT = ∠ TUV = n° (alternate segment)

Angles subtended by the diameter (or semi-circle) is 90°.

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

m° + n° + 90° = 180° (Sum of angles in a triangle)

m° + n° = 180° - 90°

m° + n° = 90°

Central angle is twice any inscribed angle subtended by the same arc.

∠ LOM = 2 x ∠ LNM

∠ LNM = $\frac{\mathrm{\angle \; LOM}}{2}$ = $\frac{\mathrm{x}}{2}$

∠ LNM + ∠ NLM + ∠ LMN = 180° (Sum of angles in a triangle)

∠ LNM = $\frac{\mathrm{x}}{2}$

∠ NLM = 74°

∠ LMN = 39°

$\frac{\mathrm{x}}{2}$ + 74 + 39 = 180

Get rid of the fraction by multiplying both sides of the equation by the denominator 2.

x + 2 x 74 + 2 x 39 = 180 x 2

x + 148 + 78 = 360

x + 226 = 360

x = 360 - 226

x = 134

∆ SOT is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ OST = ∠ OTS = 50°

∠ OST + ∠ OTS + ∠ SOT = 180°(Sum of angles in a triangle)

50° + 50° + ∠ SOT = 180°

100° + ∠ SOT = 180°

∠ SOT = 180° - 100°

∠ SOT = 80°

∠ SOT = ∠ ROP = 80° (Vertically opposite angles)

∠ ROP + ∠ OPQ = 180° (Co-interior angles)

80° + ∠ OPQ = 180°

∠ OPQ = 180° - 80°

∠ OPQ = 100°

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

Angles subtended by the same arc are the same.

∠ QSR = ∠ QPR = 52°

y = 52°

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

∠ QPR + ∠ PRQ + ∠ PQR = 180°

52° + x + 70° = 180°

x + 122° = 180°

x = 180° - 122°

x = 58°

∠ QFG + ∠ EFG = 180°

105° + ∠ EFG = 180°

∠ EFG = 180° - 105°

∠ EFG = 75°

∠ FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)

50° + m + 75° = 180°

m + 125° = 180°

m = 180° - 125°

m = 55°

n = ∠ FGS

∠ FGS = 180° - 105°(Corresponding angle to line RS)

∠ FGS = 75°

n = ∠ FGS = 75°

w + x + y + z = 180° (Sum of angles on a straight line)

But (w + x + y) = 140°

140° + z = 180°

z = 180° - 140°

z = 40°

(x + y + z) = 130°

x + y + z = 130°

x + y = 130° - z

But z = 40°

x + y = 130° - 40°

x + y = 90°

∠ PQR : ∠ QPS = 3 : 7

∠ PQR + ∠ QPS = 180° (Respective consecutive angles are supplementary)

Sum of ratios = 3 + 7 = 10

The sum of ratios represents the sum of the angles.

If 10 = 180°

3 = $\frac{\mathrm{3\; x\; 180°}}{10}$ = 54°

∴ ∠ PQR = 54°

If 10 = 180°

7 = $\frac{\mathrm{7\; x\; 180°}}{10}$ = 126°

∴ ∠ QPS = 126°

Alternatively,

∠ QPS = 180° - ∠ PQR

∠ QPS = 180° - 54° = 126°

∠ PSR = ∠ PQR = 54° (Opposite angles are equal)

∠ QSR = $\frac{\mathrm{\angle \; PSR}}{2}$ = $\frac{\mathrm{54°}}{2}$ = 27°

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43° (Vertically opposite angles)

∠ OBC = ∠ OCB = y = 43°

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43°

∠ OBC = ∠ OCB = y = 43°

∠ OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)

∠ OBC = ∠ OCB = 43°

43° + 43° + ∠ BOC = 180°

86° + ∠ BOC = 180°

∠ BOC = 180° - 86°

∠ BOC = 94°

x = $\frac{\mathrm{BOC}}{2}$ = $\frac{\mathrm{94°}}{2}$ = 47°

∆ BCD is isosceles since two sides are equal.

The base angles of isosceles triangles are the same.

∠ ECD = ∠ DBE = 35° (Base angles are equal)

∠ EDB = $\frac{1}{2}$ x ∠ CDB

∠ ECD + ∠ DBE + ∠ CDB = 180°(Sum of angles in a triangle)

35° + 35° + ∠ CDB = 180°

70° + ∠ CDB = 180°

∠ CDB = 180° - 70°

∠ CDB = 110°

∠ EDB = $\frac{1}{2}$ x 110° = 55°

∠ ABD = 90°(Angle subtended by the diameter)

∠ BAD + ∠ EDB + ∠ ABD = 180°(Sum of angles in a triangle)

m + 55° + 90° = 180°

m + 145° = 180°

m = 180° - 145°

m = 35°

∆ BCD is isosceles since two sides are equal.

The base angles of isosceles triangles are the same.

∠ ECD = ∠ DBE = 35° (Base angles are equal)

∠ ACD = 90°(Angle subtended by a diameter)

∠ ACE + ∠ ECD = 90°

n + 35° = 90°

n = 90° - 35°

n = 55°

∆ MNT is an isosceles triangle (Two sides equal).

Isosceles triangle has the base angles the equal.

∠ MNT = ∠ MTN = 32°

∠ LMN = ∠ MNT + ∠ MTN (Sum of two opposite interior angles equal to exterior angle)

∠ LMN = 32° + 32° = 64°

∠ MLN = $\frac{1}{2}$ x ∠MON

|OM| = |ON| (Radii)

∆ MON is an isosceles triangle hence the base angles are the same.

∠ OMN = ∠ ONM = 20°

∠MON = 180° - (20° + 20°) (Sum of interior angles of a triangle is 180°)

∠MON = 180° - 40°

∠MON = 140°

∠ MLN = $\frac{1}{2}$ x ∠MON = $\frac{1}{2}$ x 140° = 70°

∠ MLN + ∠ LMN + ∠ LNM = 180°(Sum of interior angles of a triangle is 180°)

64° + 70° + ∠ LNM = 180°

134° + ∠ LNM = 180°

∠ LNM = 180° - 134°

∠ LNM = 46°

Draw an imaginary third parallel line.

∠ SRH = ∠ UHR = 42° (Alternate angles)

∠ HQP = ∠ QHU = y (Alternate angles)

∠ RHQ = ∠ UHR + ∠ QHU = 65°

42° + y = 65°

y = 65° - 42°

y = 23°

∠ QRS = ∠ PQT = 40° (Corresponding angles)

∆ PQT is an isosceles triangle (Two sides equal).

Isosceles triangle has base angles equal.

∠ RPT = ∠ PQT = 40° (Base angles equal)

∠ QRT = ∠ QST = 42° (Angles subtended by arc QT)

∠ QST + ∠ RSQ = 90°

42° + ∠ RSQ = 90°

∠ RSQ = 90° - 42°

∠ RSQ = 48°

∠ STR = ∠ SQR = 180° - 124°(Angles subtended by arc SR)

∠ STR = 180° - 124° (Sum of angles on straight line is 180°)

∠ STR = 56°

Isosceles triangles (two sides equal) has base angles equal.

For ∆XYZ, ∠YXZ = ∠YZX = a

a + a + 52° = 180°(Sum of interior angles of a triangle)

2a + 52° = 180°

2a = 180° - 52°

2a = 128°

Divide both sides by 2

a = $\frac{\mathrm{128°}}{2}$ = 64°

For ∆XZT, ∠ZXT = ∠ZTX = b

∠YZX + ∠XZT = 180°(Sum of angles on a straight line)

∠YZX = a = 64°

64° + ∠XZT = 180°

∠XZT = 180° - 64°

∠XZT = 116°

∠XZT + ∠ZXT + ∠ZTX = 180°(Sum of interior angles of a triangle)

116° + b + b = 180°

2b = 180° - 116°

2b = 64°

Divide both sides by 2.

b = $\frac{\mathrm{64°}}{2}$ = 32°

∠ZTX = 32°