1. Using the COSINE Rule

2. Quick starter … answer these
cos(x) = 6 ÷ 9 = 2/3
x = cos-1(2/3)
x = 48.19°
62 + b2 = 92 b2 =
b = √45
b = 6.71cm H
9cm
9cm O b
Q 1
Q 2 x°
6cm
6cm A
cos(33°) = 11.3 ÷ y
y = 11.3 ÷ cos(33°)
y = 13.47cm A O
11.3 cm
Q 3
33° y H

3.  From the formula sheet on exams …
Seems to combine Pythagoras and cosine formula

4. The Cosine Rule
DOESN’T HAVE TO BE RIGHT-ANGLED TRIANGLE

5. Eg 1 B y
(a)
7.9 cm
(c)
116° C A
6.5 cm
(b)
y2 = – 2 x 6.5 x 7.9 x cos(116°)
y2 = – x ( …)
y = √ … = cm (2dp)

6. Eg 2 B
12 cm
(a)
6 cm
(c) y° C A
8 cm
(b)
122 = – 2 x 8 x 6 x cos(y°)
144 = 100 – 96 x cos(y°)
(-100)
44 = (– 96) x cos(y°)
44 = cos(y°) 96
÷(-96) -
y = cos-1( 44 ) = ° (2dp) 96 -

7. cos (90°) = 0 Remember … y = cos (θ) y 1 0.5 -0.5 -1 θ y 1
0.5
-0.5 -1
cos (90°) = 0
Remember … y = cos (θ)

8. 62 + b2 = 92 b2 =
b = √45
b = 6.71cm
9cm b
Q 1
6cm C
92 = b – 2 x b x 6 x cos(90°)
9cm
(a) b
81 = b – 12 x b x 0
(b)
81 = b2 + 36 B A
6cm
b2 =
b = √45
b = 6.71cm
(c)