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© T Madas
© T Madas
The Cosine Rule
© T Madas
A B
C
ab
c
a2 = b2 + c2 – 2bccosAb2 = a2 + c2 – 2accosB
c2 = a2 + b2– 2abcosCThe cosine rule
in any triangle
© T Madas
R P
Q
rp
q
p2 = r2 + q2– 2r qcosPThe cosine rule
© T Madas
x 9
45° 65°
70°
d9
8
d 2 = 92 + 82 – 2 x 8 x cos45°
The cosine rule
© T Madas
x 4
41° 68°
71°
x 4
7
x 2 = 42 + 72 – 2 x 7 x cos68°
The cosine rule
© T Madas
45° 72°
63°
x
6 cm
8 cm
Calculate the missing length in the triangle below:
UsingTheCosineRule
© T Madas
Calculate the missing length in the triangle below:
45° 72°
63°
x
6 cm
8 cmx 6x
2 = 62 + 82 – 2 x 8 x cos72° Û
x 2 = 36 + 64 – 96cos72° Û
x 2 = 100 – 96cos72° Û
x = Û- °100 96 cos 72
x ≈ 8.4 cm
© T Madas
Calculate the missing angle in the triangle below:
θ
5 cm 6 cm
8 cm
Û=26
48.5» °
25 + 28 - 2´ 5´ 8´ cos
Û=36 25 +64 - 80 cos
Û=80 cos 25 +64 - 36
Û=80 cos 53
Û=cos 5380
= ( )-1 5380cos Û
© T Madas
The Sine Rule
© T Madas
=
A B
ab
c
The Sine Rulesin A sinB
=sinC
=a b
=c
C
in any triangle
sin A sinB sinC
a b c
© T Madas
Calculate the missing length in the triangle below:
45° 72°
63°
x
6 cmUsingTheSineRule
© T Madas
°sin72 °sin45
Calculate the missing length in the triangle below:
45° 72°
63°
x
6 cm
Ûx = 6
x °sin45 = 6 °sin72 Û
x =°
°6sin72sin45
Û
8.07 cmx »
© T Madas
°sin70 °sin35
Calculate the missing length in the triangle below:
35° 75°
70°
x
8 cm
Ûx = 8
x °sin35 = 8 °sin70 Û
x =°
°8sin70sin35
Û
» 13.11 cmx
© T Madas
7 5
Calculate the missing angle in the triangle below:
35°
x
7 cm
5 cm
Ûsinx = °sin35
5 sinx = 7 °sin35 Û
sinx =°7sin35
5Û
» °53x
sinx » 0.803 Û
© T Madas
© T Madas
A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.
Base
N
4060°
A 5 B
C
30°
d
150°
© T Madas
Base
N
060°
A 5 B
C
30°
d
150°
– 40
x 5d 2= 52+42 – 2 x 4 x
cos150°
By the cosine rule on ABC
d 2= 25 + 16 cos150°
d 2 ≈ 75.64
d ≈8.7 km
4
8.7
© T Madas
A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.
Base
N
4060°
A 5 B
C
30° 150°
8.7
8.7 km
θ
© T Madas
Base
N
4060°
A 5 B
C
30° 150°
8.7θ
By the sine rule on ABC :
sinθ5
sin150°8.7
=
5sin150°8.7
sinθ =
x 55 x
0.287sinθ ≈
sin-1 (0.287)θ ≈
17°θ ≈
17°
© T Madas
Base
N
4060°
A 5
C
30° 150°
8.7
A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.
8.7 km077°
17°
B is at a bearing of 077° from the base
257°
B
77°
© T Madas
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