BAM FORM 6 – TRIGONOMETRY

 

RADIANS

Definition

1 radian is an angle subtended at the centre of the circle of the circle by an are equal in length to that radius.

i.e
                                                                                                                                                           

But θ = 1 radian

Circumference = 2

  =

  =  =  = 

  =

Radian = 180^o

Note: It is customary to omit the unit radian

 = 180^o

Exercise 1.3

1.   Convert the following into degrees

            a)       b)     c)     d)             

e) 5     f)    g)    h)

j)

2.   Convert the following into radians

a) 120^o           b) 4^o                c) -90^o            d) 15^o

e) 60^o              f) 240^o            g) -315           h) -90^o

i) 357^o            j) 300^o            k) -630^o

3. Solve for x such that 0^o ≤ x ≥ 360^o

a) Cos x = -1

b) Cos x = –

c) Sin x = –

d) Sin x = 0

e) Cos x =

f) Sin x =

g) Tan x = undefined

h) Tan x = 0

i) Tan x = -1

j) Tan x = 1

4. Evaluate sin θ if the terminal side of θ contains the given point

a) P (10, -4)               b) P (3, 4)       c) P (3, -5)     d) P (-1, 3)

e) P (-8, -13)

5. If cos θ = -2/5 and sin θ < 0, find sin θ

6. If cos θ = 5/3 and sin θ < 0, find sin θ.

                                      

Sin A = h/c             h=c sin A

Sin C = h/a             h=a sin C

Sin A / a = sin C / c

Similarly,  by drawing the altitude from A to BC.

   b sin C = c sin B

   Sin C / c= sin B / b

   Hence; sin A /a = sin B / b = sin C / c

   It is called the sine rule.

Note:

a) When two angles and one side of any triangle are known, the sine rule can be used to solve the triangle

b) When two sides and an angle opposite to any of the two given sides are known, the sine rule can be used to solve the triangle.

Example

7. In triangle ABC, B = 39⁰, C = 82⁰, a = 6.73 cm find C

Solution

From       

                =  =

But A = 180^o – (39^o + 82^o)

                        = 59^o

            =  =

            C = 6.73cm x  = 7.78 cm

Example 1.4

Find the remaining angles of the AABC in which a = 12.5 cm , c = 17.7 cm and C = 116⁰

Solution

  =  =  =

= sin A =

            A = sin^-1(0.6347)

            A = 39⁰.24

            B = 180^o – 155 = B = 24^o 36

Exercise 1.5

1.  Solve completely the triangles in which

            a) b = 6m, A= 80^o, C = 46^o

            b) A = 50^o, B = 60^o, a = 4m

            c) a = 400m, A = 31^o20, B = 70^o40

            d) A = 115^o, a = 65m, b = 32m

Cosine rule

Consider the triangle ABC

           

           By using Pythagoras theorem

            C² = (x – a) ² + h²

            C² = x2 – 2ax + a² + h²

            C² = x² + h² + a²

            C² = a² + b² – 2ax

            But  = cosC

            C² = a² + b² – 2ab Cos C

            For angle B

            b² = a² + c² – 2ac Cos B

            For angle A

            a² = b² + c² – 2bc Cos A

Note

a) When two sides and an angle between the two sides are included then the angle between the two lines of the cosine rule can be used to solve the triangle by using the cosine rule.

b) We can also solve the triangle by cosine rule if all the three sides are given

Example 1.6

Given a triangle with a = 50cm, c =60cm, and B = 100^o .Find b

Solution

b² = a² + c² – 2ac Cos B

            = (50) ² + (60) ² – 2 (50) (60) Cos 100^o

            = 3600 + 2500 – 6000 (-0.1736)

            = 7141.6 cm

Example 1.7

Given a = 8cm, b = 5cm, and c = 4cm, find the smallest and largest angles.

Solution

i) To find A we use

            a² = b² + c² – 2bc Cos A

            64 = 25 + 16 – 40 Cos A

            40cos A = 41 – 64

            40cos A = -23

            Cos A =  = -0.5750

A = 125^o10

ii) To find C we use

            c² = a² + b² – 2abCos C

            16 = 8² + 25 – 2 (8) (5) Cos C
16 = 64 + 25 – 2(8)(5)Cos C

            80cos C = 89 – 16

                        Cos C =  = 0.9125

                                    C = 24^o10

Trigonometric identities

Figure 8.15 show the right angled triangle with sides a, b, c and angles A, B, C

By the use of Pythagoras theorem

            a² + b² = c²

Divide throughout by c²

But    = cos θ,  = sin θ

        Cos² θ + sin ² θ = 1

Exercise 1.8

1. Use the cosine rule to solve the following triangles

a) A = 60^o, b = 3cm, c = 4cm

b) A = 100^o, b = 8cm, c = 10cm

c) A = 50^o, b = 5cm, c = 8cm

d) a = 9cm, b = 10, C = 120^o

e) a = 5, b = 10, c = 120⁰

f) a = 10cm, b = 20cm, c = 120cm

2. Find the largest and smallest angles of triangle whose side area a= 5cm b = 10cm and c = 12cm

3. Two people are on opposite sides of a hill and are 600m apart. If the angles of elevation from these people to the top of the hill are 19^o and 21^o, how high is the hill?