1.  1 1 1 O Radians In a circle of radius 1 unit, the angle  subtended at the centre of the circle by the arc of length 1 unit is called 1 radian, written as 1 rad. The circumference of the circle is given by C = 2r. When the radius is 1 unit then C = 2 radian. 2 radians = 360° :  radians = 180° ( 1 rad  57.3°) Converting between degrees and radians To convert degrees to radians, multiply by To convert radians to degrees, multiply by

2. Examples Write the following angles into radians, leaving your answers as a multiple of . (a) 60° (b) 45° (c) 120° (d) 72° (e) 270° (a) 60° = 60  =  (b) 45° = 45  = ¼  (c) 120° = 120  =  (d) 72° = 72  =  (e) 270° = 270  = 1.5 

3. Examples Write the following angles into degrees. (a) 2 rad (b) /5 rad (c) 0.3 rad (d) 4.2 rad (e) 0.04 rad. (a) 2 = 2  = 360º (b) /5 = /5 = 36º (c) 0.3 rad = 0.3  = 17.2º (d) 4.2 rad = 4.2  = 240.6º (e) 0.04 rad = 0.04  = 2.29º

4.  L r O Length of arc The length of a circular arc with radius r and angle  rad is L= r. Example The sector of a circle of radius 5 cm subtends an angle of ¼  rad at the centre, find the length of the arc of the sector The length, L = r = 5  ¼  = 1.25  = 3.93 cm. Example L = 10 cm, r = 5.5 cm find . L = r   10 = 5.5    = 1.82 rad or  =104º

5.  L r O Area of sector The area of a circular sector with radius r and angle  rad is A = ½ r2. Example The sector of a circle of radius 5 cm subtends an angle of ¼  rad at the centre, find the area of the sector of the circle. The area, A = ½ r2 = 0.5  52 ¼  = 9.82 cm2 . Example A = 25 cm2,  = 0.6 rad, find r and L A = ½ r2 = 0.5 x r2x 0.6 = 25  r = 9.13 cm L = 5.48 cm

6. b C a 8 cm 70º 12 cm Area of a triangle When we know two sides and the angle between them, we can use a formula to calculate the area of a triangle. Area = Area = = 0.5 12  8  sin 70º = 45.1 cm2

7. Area of segment Area of segment S = Area of sector – area triangle Area S = Area S = O   is in radians r r s

8. Circle