60º
DESCRIPTION
60º. 2. 2. 2. 60º. 60º. 2. 60º. Exact Values. Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values. 30º. 3. 1. This triangle will provide exact values for sin, cos and tan 30º and 60º. 30º. 2. 3. 60º. 1. 1. 0. 1. 0. ∞. 0. - PowerPoint PPT PresentationTRANSCRIPT
2222
22
60º60º
60º60º60º60º11
60º60º
2230º30º
33
This triangle will provide exact values for This triangle will provide exact values for
sin, cos and tan 30º and 60ºsin, cos and tan 30º and 60º
Exact ValuesExact ValuesSome special values of Sin, Cos and Tan are useful left as Some special values of Sin, Cos and Tan are useful left as fractions, We call these fractions, We call these exact valuesexact values
xx 0º0º 30º30º 45º45º 60º60º 90º90º
Sin xºSin xº
Cos xºCos xº
Tan xºTan xº
½
½½
3
23
23
31
1160º60º
2230º30º33
Exact ValuesExact Values
0
1
0
1
0
∞∞
Exact ValuesExact Values
11 1145º45º
45º45º
22
Consider the square with sides 1 unitConsider the square with sides 1 unit
1111
We are now in a position to calculate We are now in a position to calculate exact values for sin, cos and tan of 45exact values for sin, cos and tan of 45oo
Exact ValuesExact Values11
45º45º
45º45º
22
11
Tan xºTan xº
Cos xºCos xº
Sin xºSin xº
90º90º60º60º45º45º30º30º0º0ºxx
00
00
11
11
00
21
21
23
23
33
1
21
21
11
Apr 20, 2023Apr 20, 2023
Naming the sides of a TriangleNaming the sides of a Triangle
AA
BB
CC
aa
bb
cc
Apr 20, 2023Apr 20, 2023
Area of ANY TriangleArea of ANY Triangle
AA
BB
CC
aa
bb
cc
The area of ANY triangle can be found The area of ANY triangle can be found by the following formula.by the following formula.
Another versionAnother version
Another versionAnother version
Key feature Key feature
To find the areaTo find the areayou need to knowyou need to know
2 sides and the angle 2 sides and the angle in between (SAS)in between (SAS)
AbcArea sin21
BacArea sin21
CabArea sin21
If you know A, b and cIf you know A, b and cIf you know B, a and cIf you know B, a and cIf you know C, a and bIf you know C, a and b
Remember:Remember:
Sides use lower case Sides use lower case letters of anglesletters of angles
Area of ANY TriangleArea of ANY Triangle
AA
BB
CC
AA
20cm20cmBB
25cm25cm
CCcc
Example : Find the area of the triangle.Example : Find the area of the triangle.
The version we use isThe version we use is
3030oo
120 25 sin 30
2oArea
210 25 0.5 125Area cm
CabArea sin21
aa
bb
Area of ANY TriangleArea of ANY Triangle
DD
EE
FF
10cm10cm
8cm8cm
Example : Find the area of the triangle.Example : Find the area of the triangle.
sin1
Area= df E2
The version we use isThe version we use is
6060oo
18 10 sin 60
2oArea
240 0.866 34.64Area cm
d
e
f
What Goes In The Box ?What Goes In The Box ?
Calculate the areas of the triangles Calculate the areas of the triangles below:below:
(1)
23o
15cm
12.6cm
(2)
71o
5.7m
6.2m
A =36.9cm2
A =16.7m2
Key Key feature feature
Remember Remember (SAS)(SAS)
Apr 20, 2023Apr 20, 2023
17 cm17 cm
95 95 cmcm22
12 cm12 cm
PP
RR
Finding the Angle given the areaFinding the Angle given the area
Find the size of angle PFind the size of angle P
pp
rr
Area = Area = ½½qrqrsinPsinP
95 = 95 = ½ ½ 1717 × × 1212 sinPsinP
95 = 95 = 102102 sinPsinP
sinP sinP = 95 ÷ 102 = 95 ÷ 102 = 0∙931 = 0∙931
P = sinP = sin-1-1 0∙931 = 68∙6° 0∙931 = 68∙6°
The side opposite angle The side opposite angle AA is labelled is labelled aa
The Sine RuleThe Sine Rule
Bsinb
Asin
a
AA
BB
CC
aa
bb
cc
The side opposite angle The side opposite angle BB is labelled is labelled bbThe side opposite angle The side opposite angle CC is labelled is labelled cc
Csinc
The RuleThe Rule
Calculating Sides Using The Sine RuleCalculating Sides Using The Sine Rule
Find the length of Find the length of xx in in this triangle.this triangle.
ox
41sin
o34sin
10 Now cross Now cross multiply.multiply.
oox 41sin1034sin
o
ox
34sin
41sin10
mx 74.11559.0
656.010
Example 1Example 1
PP3434oo
4141oo
xx10m10m
RR
Rr
Pp
sinsinsin
1010
sin 34sin 34°°
xx
sin 41sin 41°°
Find the length of x in this triangle.
ox
133sin
o37sin
10
oox 133sin1037sin
o
ox
37sin
133sin10
602.0731.010
x = 12.14m= 12.14m
Example 210m133o
37o
xD
E
F
Ff
Ee
Dd
sinsinsin
1010
sin 37sin 37°°
xx
sin 133sin 133°°
Cross Cross multiplymultiply
The balloon is anchored to the ground as shown in the diagram.
o70sin
25
o
x
35sin
Problem
C
c
B
b
A
a
sinsinsin
xx
sin 35sin 35°°
2525
sin 70sin 70°°
25 m25 m
7575°°xx m
7070°°
Calculate the distance between the anchor points.
A
B Caa
bbcc
3535°°
sin 75sin 75°°
oox 35sin2570sin
oox 70sin35sin25
mx 315
Cross Cross multiplymultiply
Find the unknown side in each of the triangles below:Find the unknown side in each of the triangles below:
(1) 12cm
72o
32o
x(2)
93o
y47o
16mm
(3)
87o
89m
35o
a (4) 143o
g12o
17m
x = 6.7cm y = 21.8mm
a = 51.12m51.12m g = 49.21m
Calculating Angles Using The Sine RuleCalculating Angles Using The Sine Rule
Example 1.
Find the angle Find the angle aaoo
oasin
45o23sin
38
ooa 23sin45sin38
38
23sin45sin
ooa = 0.463
ooa 6.27463.0sin 1
ao
45m
23o
38m
Z
Y
X
Zz
Yy
Xx
sinsinsin
4545
sin sin aaºº
3838
sin 23sin 23ºº
Cross Cross multiplymultiply
Use sinUse sin-1-1
Example 2.
143o
75m
38m
boFind the size of Find the size of the angle bthe angle boo
obsin
38
oob 143sin38sin75
o143sin
75
75
143sin38sin
oob = 0.305
oob 8.17305.0sin 1
Calculate the unknown angle in the following:Calculate the unknown angle in the following:
(1)
14.5m
8.9m
ao
100o(2)
14.7cm
bo
14o
12.9cm
(3)93o
64mm
co
49mm
aaoo = 37.2 = 37.2oo bbo o = 16= 16oo
c = 49.9c = 49.9oo
Sine Rule
Basic Examples
In the examples which follow use the sine rule to find the required side or angle.
asin A
bsin B
csinCB
5.8
C74° 34°A
Find AB
5.7
9.6
76°
C
B
A
Find angle B
DD
4.64.6
EE80°80°
3.23.2
FF Find angle FFind angle F
In the examples which follow use the sine rule to find the required side or angle.
asin A
bsin B
csinC
13.8
127°127°
36°36°
P
R
Q
Find PQ
M
12
N
86°86° 7
TFind angle NFind angle N
A
8
B
C
72° 72°
Find ABFind AB
RR60°60°
SS
78°78°
3.83.8
TTFind RTFind RT
II
27°27°
130°130°4.14.1
GG
HH
Find GIFind GI
AA
99
BB
75°75°
32°32°
CC
Find BCFind BC
88
39°39°
1010
YY
XX
ZZFind angle ZFind angle Z
BB
70°70°
CC
99
1313
AAFind angle AFind angle A
Sine Rule (Bearings)Sine Rule (Bearings)
TrigonometryTrigonometry
In the examples which follow use the sine ruleIn the examples which follow use the sine rule to find the required side or angle.to find the required side or angle.
asin A
bsin B
csinC
Consider two radar stations Alpha and Beta. Consider two radar stations Alpha and Beta. Alpha is 140 miles west of Beta.Alpha is 140 miles west of Beta. The bearing of an aero plane from Alpha is 032° The bearing of an aero plane from Alpha is 032° and from Beta it is 316°. and from Beta it is 316°. How far is the aeroplane from each of the radar stations? How far is the aeroplane from each of the radar stations?
1.1.
2.2. Two ports Dundee and Stonehaven are 54 miles apart with Two ports Dundee and Stonehaven are 54 miles apart with Dundee approximately due south of Stonehaven. Dundee approximately due south of Stonehaven. The bearing of a ship at sea from Stonehaven is 098° The bearing of a ship at sea from Stonehaven is 098° and from Dundee it is 048°.and from Dundee it is 048°.
How far is the ship from Dundee? How far is the ship from Dundee?
3.3. A ship is sailing north. At noon its bearing from a lighthouse A ship is sailing north. At noon its bearing from a lighthouse is 240°. Five hours later the ship is 84km further north and is 240°. Five hours later the ship is 84km further north and its new bearing from the lighthouse is 290°. its new bearing from the lighthouse is 290°. How far is the ship from the lighthouse now? How far is the ship from the lighthouse now?
4. Two ports P and Q are 35 miles apart with P due east of Q. The bearings of a ship from P and Q are 190° and 126° respectively.
a) What is the bearing of port Q from the ship?
b) How far is the ship from port P
5. The bearing of an aeroplane from Aberdeen is 068° and from Dundee it is 030°. Dundee is 69 miles south of Aberdeen. How far is the aero plane from Aberdeen?
6.6. Two oil rigs are 80 miles apart with rigTwo oil rigs are 80 miles apart with rig B being 80 miles east of rig A. B being 80 miles east of rig A.
The bearing of a ship from rig B is 194° The bearing of a ship from rig B is 194° and from A the bearing is 140°. and from A the bearing is 140°. Find the distance of the ship from rig A. Find the distance of the ship from rig A.
7.7. At noon a ship lies on a bearing 070° from Dundee. At noon a ship lies on a bearing 070° from Dundee. The ship sails a distance of 45 miles south and at 3pm its The ship sails a distance of 45 miles south and at 3pm its new bearing from Dundee is 130°. new bearing from Dundee is 130°.
How far is the ship from Dundee now? How far is the ship from Dundee now?
If the ship maintains the same speed, If the ship maintains the same speed, how long will it take to return to Dundee?how long will it take to return to Dundee?
8.8. An aeroplane leaves Leuchers airport and flies 80 miles north.An aeroplane leaves Leuchers airport and flies 80 miles north. It then changes direction and flies 120 miles east. It then changes direction and flies 120 miles east. The plane now turns onto a bearing of 126° The plane now turns onto a bearing of 126°
and flies a further 164 miles. and flies a further 164 miles.
Calculate the bearing and distance of Calculate the bearing and distance of the plane from Leuchers. the plane from Leuchers.
QuestionQuestion SolutionSolution
11 122.4 miles ; 103.8 miles122.4 miles ; 103.8 miles
22 69.8 miles69.8 miles
33 95.0 kilometers95.0 kilometers
44 Bearing 306° ; 22.9 milesBearing 306° ; 22.9 miles
55 56.0 miles56.0 miles
66 114.9 miles114.9 miles
77 48.8 miles ; 3 hours 15 minutes48.8 miles ; 3 hours 15 minutes
88 236.2 miles ; bearing 269°236.2 miles ; bearing 269°
Note: the last question requires the cosine rule as well as the sine rule.
C
B
AApr 20, 2023Apr 20, 2023
Cosine RuleCosine Rule
a
b
c
The Cosine Rule can be used with ANY triangle The Cosine Rule can be used with ANY triangle
as long as we have been as long as we have been givengiven enough information enough information.
Abccba cos2222
Bcaacb cos2222
c2 a2 b2 2abcos C
Given Given angle Aangle AGiven Given angle Bangle BGiven Given angle Cangle C
Using The Cosine RuleUsing The Cosine RuleExample 1 : Find the unknown side in the triangle below:Example 1 : Find the unknown side in the triangle below:
Identify sides Identify sides a, b, ca, b, c and and angle angle AAoo
aa = = xx bb = = 55 cc = =1212 AAo o == 4343ooWrite down the Cosine Rule Write down the Cosine Rule
for for aa
Substitute valuesSubstitute valuesxx22 = = 5522 ++ 121222 - 2 - 2 xx 5 5 xx 12 cos 43 12 cos 43oo
xx22 = = 81.2881.28 Square root to find “Square root to find “xx”.”.
xx = 9.02m = 9.02m
xx5m5m
12m12m
4343oo
AA BB
CC
aa22 = = bb22 + + cc22 - 2 - 2bcbc cos cos AA
pp22 = = qq22 + + rr22 – 2 – 2pqpq cos P cos P
Example 2Example 2 : :
Find the length of side QRFind the length of side QR
Identify the sides and angle.Identify the sides and angle.p = yp = y rr = 12.2 = 12.2 qq = 17.5 = 17.5 PP = 137= 137oo
Write down Cosine RuleWrite down Cosine Rule for for pp
yy22 = 12.2 = 12.222 + 17.5 + 17.522 – 2 – 2 xx 12.2 12.2 xx 17.5 17.5 xx cos 137 cos 137oo
yy22 = 767.227 = 767.227
yy = 27.7m = 27.7m
Using The Cosine RuleUsing The Cosine Rule
137137oo 17.5 m17.5 m12.2 m12.2 m
yy
PP
QQ RR
SubstituteSubstitute
Find the length of the unknown side in the triangles:Find the length of the unknown side in the triangles:
(1)(1)7878oo
43cm43cm
31cm31cmpp
AA
BB
CC
a
b
c
a2 = b2 + c2 – 2bc cosA
p2 = 432 + 312 – 2 × 43 × 31 × cos78°
p2 = 2255∙7
p = 47∙5 cm
Find the length of the unknown side in the triangles:Find the length of the unknown side in the triangles:
(2)(2)
8m8m
55∙∙2m2m
3838oo
mmm = 5m = 5∙∙05m05m
112112ºº17 mm17 mm28 mm28 mm
kk
k = 37k = 37∙∙8 mm8 mm
(3)(3)
Cosine RuleCosine Rule Basic ExamplesBasic Examples 1. Finding a side. 1. Finding a side.
a2 b2 c2 2bc cos A
5.6cm5.6cm
39°39°
CC
BB4.7cm4.7cmAA
Find aFind a
2.4cm2.4cm36°36°
CC
BB
2.72.7cmcm
AA Find bFind b
104°104°7.4cm7.4cm
CC
AA
3.8cm3.8cm
Find aFind a BB
CC Find cFind c
6cm6cm
BB
32°32°
AA
6cm6cm
CC
1.1.2.2.
3.3.4.4.
BB
6.6cm6.6cm 44°44°
CC
BB
8.7cm8.7cm
AAFind cFind c
8cm8cm
67°67°
BB AAFind a.Find a.
6cm6cm
CC
PP
44
RR
44
134°134°
Find qFind q
3cm3cm71°71°
NN
PP
5cm5cm
MM
Find mFind m40°40°11.4cm11.4cm
TT
UU
8.8cm8.8cmFind vFind v
VV
Find aFind a
4.5cm4.5cm
22°22°
AA
CC
1.7cm1.7cm 5.5.
6.6. 7.7.
9.9.
1010..
8.8.
Finding Angles Finding Angles Using The Cosine RuleUsing The Cosine Rule
The Cosine Rule formula can be rearranged to The Cosine Rule formula can be rearranged to allow us to find the size of an angleallow us to find the size of an angle
bcacb
A2
cos222
This formula is cyclic, depending on the angle This formula is cyclic, depending on the angle to be foundto be found
acbca
B2
cos222
ab
cbaC
2cos
222
Label and identify angles and sides
D = xo d = 11 e = 9 f = 16
Substitute values into the formula.
cos x = 0.75Use cos-1 0∙75 to find x
x = 41∙4o
Example 1 : Calculate the
unknown angle, xo .
Finding Angles Using The Cosine Rule
D E
F
Write the formula for cos Dcos D =e 2 + f 2 - d 2
2ef
cos x =92 + 162 - 112
2 x 9 x 16
de
f
Example 2: Find the unknown angle in the triangle:Example 2: Find the unknown angle in the triangle:
Write down the formula for cos BWrite down the formula for cos B
Label and identify the Label and identify the sides and angle.sides and angle.
B = yB = yoo
aa = 13 = 13bb = 26 = 26
cc = 15 = 15
The negative tells you the angle is obtuse.The negative tells you the angle is obtuse.
yy = 136.3 = 136.3oo
AA
BB
CC
cos B =cos B = a a 22 ++ c c 22 -- b b 22
22acac
Substitute valuesSubstitute values
cos cos yy == - 0- 0∙∙723723
cos cos yy = =131322 ++ 151522 -- 262622
2 2 x x 13 13 x x 1515
Use cosUse cos-1-1 -0 -0∙∙723 to find 723 to find yy
Calculate the unknown angles in the triangles below:Calculate the unknown angles in the triangles below:
(2)(2)
10m10m
7m7m5m5m
bboo
aaoo
(1)(1)
12.7cm12.7cm
7.9cm7.9cm 8.3cm8.3cm
bboo =111.8 =111.8oo
aaoo = 37.3 = 37.3oo
A
B
C
B
C D
Basic ExamplesBasic Examples 2. Finding an angle. 2. Finding an angle.
cos A b2 c2 a2
2bc
5.6 5.6 cmcm
4.1 4.1 cmcm
CC
BB4.7cm4.7cmAA
Find AFind A
5.25.2cmcm
3.23.2cmcm
CC
AA
3.6cm3.6cmBB
CC
6cm6cm
BB
5cm5cm
AA
6cm6cm
2.4cm2.4cm
1.9cm1.9cm
CC
BB
2.72.7cmcm
AA
3.6cm3.6cm
4.7cm4.7cmAA
4cm4cm
CC 10.5cm10.5cm
7.47.4cmcm
CC
AA
5.85.8cmcm
BB
Find BFind B
Find CFind C
Find BFind BFind CFind C Find AFind A
BB
1.1. 2.2.3.3.
4.4.5.5. 6.6.
MM
4.5cm4.5cm
2.82.8cmcm
LL
4.5cm4.5cmKK
6.6cm6.6cm
6.9cm6.9cm
ZZ
XX
8.7cm8.7cm
YY
8cm8cm7cm7cm
BB AA
Find the Find the largest angle.largest angle.
6cm6cm
CC
PP
44
RR
44
7.6cm7.6cm
Find all the anglesFind all the angles
3cm3cm
4.5cm4.5cm
NN
5cm5cm
MM
Find the Find the smallest anglesmallest angle
9.6cm
11.9cmT
U
8.8cm
Find the largest angle
V
7.7.
Find LFind L
Find XFind X
8.8. 9.9.
10.10.
11.11.12.12.
1.1. Do you know the length of ALL the sides? Do you know the length of ALL the sides?
Cosine Rule or Sine RuleCosine Rule or Sine Rule
How to determine which rule to useHow to determine which rule to use
2.2. Do you know 2 sides and the angle in between? Do you know 2 sides and the angle in between?
SASSASOROR
If YES to either of the questions then Cosine RuleIf YES to either of the questions then Cosine Rule
Otherwise use the Sine RuleOtherwise use the Sine Rule
Two questionsTwo questions
Calculate the size of Calculate the size of xx in each of these diagrams in each of these diagrams
The Sine Rule a b cSinA SinB SinC
Application Problems
25o
15 m AD
The angle of elevation of the top of a building
measured from point A is 25o. At point D which is
15m closer to the building, the angle of elevation is
35o Calculate the height of the building.
T
B
Angle TDA =
145o
Angle DTA =
10o
o o
1525 10
TDSin Sin
o15 2536.5
10Sin
TD mSin
35o
36.5
o3536.5TB
Sin
o36.5 25 0. 93TB Sin m
180 – 35 = 145o
180 – 170 = 10o
The Sine Rule a b cSinA SinB SinC
A
The angle of elevation of the top of a column measured from point A, is 20o. The angle of elevation of the top of the statue is 25o. Find the height of the statue when the measurements are taken 50 m from its base
50 m
Angle BCA =
70o
Angle ACT = Angle ATC =
110o
65o
53.21 m
B
T
C
180 – 110 = 70o 180 – 70 = 110o 180 – 115 = 65o
20o25o
5o
oo 65sin
21.53
5sin
TC
o
o
65sin
5sin21.53 TC
=5.1 m=5.1 m
AC50
20cos o
o20cos
50 AC
m21.53
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles.
(a) Make a sketch of the journey.
(b) Find the bearing of the lighthouse from the harbour. (nearest degree)
The Cosine Rule
Application Problems
2 2 2
2b c a
CosAbc
H40 miles
24 miles
B
L
57 miles
A
o20.4A
Bearing Bearing = 90 – 20 = = 90 – 20 = 070070°°
2020°°
NN
40572244057
cos222
A
2 2 2
2b c a
CosAbc
The Cosine Rule a2 = b2 + c2 – 2bcCosA
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles.
Find the bearing of Q from point P.
P
670 miles
W
530 miles
Not to Scale
Q
520 miles6705302520670530
cos222
P
PP = 48.7 = 48.7° (49°)° (49°)
Bearing Bearing = 180 + = 180 + 49 = 49 = 229°229°
COSINE RULE
1. Find the length of the third side of triangle ABC when
i) b = 2, c = 5, A = 60°
ii) a = 2, b = 5, A = 65°
iii) a = 2, c = 5, B = 115°
iv) b = 6, c = 8, A = 50°
2. Find QR in ∆PQR in which PR = 4, PQ = 3, P = 18°
3. Find XY in ∆ XYZ in which YZ = 25, XZ = 30, Z = 162°
4. In ∆ ABC, a = 7, b = 4, C = 53°. Calculate c.
5. In ∆ ABC, b = 4·2, c = 6·5, A = 24°. Calculate a.
6. In ∆ ABC, a = 1·64, c = 1·64, B = 110°. Calculate b.
7. In ∆ ABC, a = 18·5, b = 22·6, C = 72·3°. Calculate c.
8. In ∆ABC, a = 100, b = 120, C = 15°. Calculate c.
9. In ∆ABC, b = 80, c = 100, A = 123°. Calculate a.
10. A town B is 20 km due north of town A and a town C is 15 km north-west of A. Calculate the distance between B and C.
11. Two ships leave port together. One sails on a course of 045° at 9 km/h and the other on a course of 090° at 12 km/h.
After 2h 30 min, how far apart will they be?
12. From a point O, the point P is 3 km distant on a bearing of 040° and the point Q is 5 km distant on a bearing of 123°.
What is the distance between P and Q ?
COSINE RULE (1) Solutions
1. Find the length of the third side of triangle ABC when
i) a2 = 22 + 52 225cos60° = 4 + 25 200·5 = 19
a = 4·36
ii) c2 = 22 + 52 225cos65° = 4 + 25 200·4226
= 20·54763 c = 4·533
iii) b2 = 22 + 52 225cos115° = 4 + 25
225(0·42262 )
= 29 + 8·45237 = 37·45237
b = 6·12
iv) a2 = 62 + 82 268cos50° = 36 + 64 960·64279
= 100 661·7076 = 38·29239
a = 6·188
2. QR2 = 42 + 32 243cos18° = 16 + 9 240·951057 = 2·17464
QR = 1·4747
3. XY2 = 252 + 302 22530cos162° = 625 + 900 1500(0·951057 )
= 1525 + 1426·5848
= 2951·5848
XY = 54·33
4. c2 = 72 + 42 274cos53°
= 31·2984
c = 5·5945
5. a2 = 4·22 + 6·52 24·26·5cos24°
= 17·64 + 42·25 554·60·91355 = 10·0104
a = 3·164
6. b2 = 1·642 + 1·642 21·641·64cos110°
= 22·6896 22·6896(0·34202 ) =
7·2190
b = 2·6868
7. c2 = 18·52 + 22·62 218·522·6cos72·3° =
598·778
c = 24·50
8. c2 = 1002 + 1202 2100120cos15° =
1217·780
c = 34·90
9. a2 = 802 + 1002 280100cos123° = 25114·225 a = 158·47
10. BC2 = 152 + 202 21520cos45° = 225 + 400 6000·7071 = 200·7359
BC = 14·17 km
45°
West
15 km
20 km
North
C
A
B
11. QR2 = 302 + 22·52 23022·5cos45°
= 900 + 506·25 954·59 = 451·6558
QR = 21·25 km NorthQ
P
22·5 km
30 km
45°R
12. PQ2 = 32 + 52 235cos83° = 9 + 25 300·12187
= 30·344PQ = 5·51 km
Cosine Rule
Bearings problems
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the
distance between the two ships.
1. Ship1 [ 74km, 053° ] ; Ship2 [ 104km, 112° ]
Port
Ship1
Ship2
North
Abccba cos2222
5353°°
112112°°
5959°°
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between
the two ships.
2. Ship1 [ 56km,021° ] ; Ship2 [ 66km,090° ]
Port
Ship1
Ship2
North
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between
the two ships.
3. Ship1 [ 83km,060° ] ; Ship2 [ 80km,159° ]
Port
Ship1
Ship2
North
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between
the two ships.
4. Ship1 [ 50km,090° ] ; Ship2 [ 62km,142° ]
Port Ship1
Ship2
North
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between
the two ships.
5. Ship1 [ 80km,146° ] ; Ship2 [ 70km,190° ]
Port
Ship1Ship2
North
Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between
the two ships.
6. Ship1 [ 47km,180° ] ; Ship2 [ 54km,235° ]
Port
Ship1
Ship2
North
Bearings Problems
1. Dundee is 84 km due south of Aberdeen. A ship at sea is on a bearing of 078° from Aberdeen and 048° from Dundee. How far is the ship from Dundee and from Aberdeen ?
2. An aeroplane is 240 km from an airport on a bearing of 100° while a helicopter is 165 km from the airport on a bearing of 212°. How far apart are the aircraft ?
Bearings Problems
3. A ship sails 80 km on a bearing of 060° from its home port. It then sails 93 km on a bearing of 134°. How far is it now from its home port ?
4. Glasgow airport is 73 km from Edinburgh airport and lies to the west of Edinburgh airport. The bearing of an aeroplane from Glasgow airport is 040° while its bearing from Edinburgh airport is 300°. How far is the aeroplane from each airport ?
Bearings Problems
5. A ship sails 74 km south from its port to a lighthouse. It then sails 85 km on a bearing of 160°. How far is the ship from the port ?
6. From a port P, ship A is 144 km distant on a bearing of 036° and ship B is 97 km distant on a bearing of 114°. What is the distance between the two ships ?
Bearings Problems
7. A ship sails 93 km on a bearing 054° and then another 108 km on a bearing of 110°. How far is it now from its starting point ?
8. Two radar stations Alpha and Beta pick up signals from an incoming aircraft. Alpha is 40 km east of Beta and picks up the signals on a bearing of 300°. Beta picks up the signals on a bearing of 070°. How far is the aircraft from Alpha and from Beta ?
Bearings Problems
9. Aeroplane P is 200 km from an airport on a bearing of 208° while Aeroplane P is 200 km from an airport on a bearing of 208° while aeroplane Q is 170 km from the same airport on a bearing of 094°. aeroplane Q is 170 km from the same airport on a bearing of 094°.
How far apart are the two aeroplanes ?How far apart are the two aeroplanes ?
10. The bearings and distances from Aberdeen of three oil platforms are :10. The bearings and distances from Aberdeen of three oil platforms are :
a) 028° 116 kma) 028° 116 km b) 081° 104 kmb) 081° 104 km c) 138° 97 kmc) 138° 97 km
A supply boat leaves Aberdeen, visits each platform and thenA supply boat leaves Aberdeen, visits each platform and then returns to Aberdeen. returns to Aberdeen.
Find the total length of the journey.Find the total length of the journey.