Sine and cosine Rule-examcheatcode

Extracted Content: Sine and cosine Rule-examcheatcode

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

This is chapter 1 of the book, Exam Cheat Code: Maths 4. The book is
designed to prepare students for Zimsec/Cambridge Exams. The book
come with the following:

** Step by step worked examples.

“* Detailed answers.

“+ Exam type Questions.

<em>“</em> Your Name, Nickname and Surname.

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Sine Rule

The Sine Rule states that for any triangle (not just right-angled triangles), the ratio of
the length of a side to the sine of its opposite angle is constant.

A

B 5 C

The ratio can be written as:

When looking for sides.

a b c

sinA sinB sin€
When looking for angles.

sinA _ sinB _ sinC

a b c

where:
a, b, c = sides of the triangle

A, B, C = angles opposite sides a, b, c
When to Use the Sine Rule

e« When you know two angles and one side (AAS or ASA cases).

e« When you know two sides and an angle not included between them (SSA
case).

e Can apply to all triangles right-angled and non-right angled. The Sine Rule is
particularly useful for non-right-angled triangles.

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

Given triangle ABC, with A= 30°, B= 45° and a= 7cm, find b

A
30°
b
45°
B 7cm Cc
A
30°
b
45°
B 7cm C

We are looking for a side b, we have information for sides and angles of A and B, we
have nothing on C.

When looking for sides.

a b

sinA sinB
7 _ Bb

sin30° sin 45°

: 7 Db :
sin45° x —— = — x sin45° (multiply both sides by sin 45° to isolate b)
sin 30° sin 45°
: 7 Db :
sin45° x— =— x sin 45% (sin 45° cancels out on the RHS)
sin 30° sin452
sin 45°x7 —p
sin 30°
7 sin 45°
“ =b (use calculator in degrees mode)
sin 30°
9.90cm=b
- b= 9.90cm (to one decimal place)
Example 2

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

Given triangle ABC with sides a =8cm, b=10cm and angle BAC = 50°, find B.

A
50°
10cm
B 8cm C
A
50°
10cm
B 8cm C

We are looking for angle B, we have information for sides and angles of A and B, we
have nothing on C.

When looking for angles.

sinA _ sinB
a b

sin50° _ sinB

8 10
nos
10 x Sino — _ x 10 (multiply both sides by 10 to isolate sinB)
nos
10 x Sino — _ x 10 (10 cancels out on the RHS)
; 0
10 x =" = sinB
an 500
— 50 =sinB
0.9576 = sinB

sin71 0.9576 = B

73.2° =B

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

- B= 73.2° (to one decimal place)

Exercise la (30 marks] Answers

Leave lengths to 3 significant figures and angles to 1 decimal place.

1. Given that YXZ = 64°, XZY = 50°,x = 11.6cm. Find z

X
649
Zz
Y OS
[3]

11.6cm
2. Given that PQR = 47°, q=6cm, p=8cm. Find angle P.

P
L>
6cm
x IN
Q 8cm R [3]

3. Given that BAC = 62°, ACB = 30°, b=10cm. Find a

A
62°
10cm
B 30° C
a [3]
4. Given that ABC = 48°12°, c=9.7cm and b=8.8cm. Find angle C.
A

C BI
5. In AABC, A = 40°, B = 60°, a = 15 cm. Find the length of side b. [3]

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

6. In APQR, p = 12 cm, Q = 45°36’, R = 60°12". Calculate the length of side q. [3]
7. In AMNO, m = 8 cm, N = 53°, O = 47°. Calculate the length of side n. [3]

8. In AXYZ, x = 10 cm, y = 12 cm, X=49° cm. Find angle Y. [3]

9. In AEFG, e = 18 cm, f = 12 cm, E = 67°19". Calculate angle F. [3]

10. In AJKL, j = 20 cm, k = 24 cm, J = 45°. Find angle K. [3]

Cosine Rule

The Cosine Rule is useful for finding the lengths of sides or angles in any triangle, not
just right-angled triangles. The Cosine Rule is particularly useful for non-right-angled
triangles.

A

B a C

Finding sides:

a? = b? +c? —2becosA
b? = a? +c? — 2accosB
c? = a? + b* — 2abcosC

Finding Angles:

b?+c?-a?
cos A = ———
2be
a? +c2—b?
cos B = ———
2ac
24 p22
a<em>+b</em>-c
cos C = ———
2ab
where:

a, b, c = sides of the triangle

A, B, C = angles opposite sides a, b, c
When to Use the Cosine Rule

e« When you know two sides and the included angle (SAS case).
e« When you know all three sides and need to find an angle (SSS case).

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

Example 3

Given that ACB = 60°,b=8cm and a=7cm. Find c

A
c 8cm
60°
B 7cm Cc

c? = a? + b? — 2abcosC
c?=77 +87-2x7xX8x cos 60°
c? =49 + 64-112 x cos60°
c? = 113 — 112 cos 60° (use a calculator in degree mode)
c? =57
Vc? = V57
c=7.55cem (to 3 significant figures)
Example 4

Given that triangle ABC sides AC=6cm, AB=5cm and BC=7cm. Find all its angles.

A

B 7cm C

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

>

L\ NSN
B 7cm C

Finding angle, A

b?+c?-a?

cosA =
2be

_ 6452-7?
~~ 2x6x5

__ 36+25-49
~ 60

_ 12
cosA = é0

A=cos"! (2)
A=78.5°

Finding angle B

a<em>+c?—b</em>

cos B =
2ac

_ 7?+5?-6?
2x7x5

__ 49+25-36
70

_ 38
cosB = i

B=cos"! (=)

B=57.1°
Finding angle C
Angles in triangle add up to 180°, since we now have two of the angles so to find the
third, we add the two and subtract from 180°

C = 180° — (78.5° +57.1°)
= 180° — 135.6°
= 44,4°

Exercise {b (30 marks] [Answers]

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

Leave lengths to 3 significant figures and angles to 1 decimal place.

1. Find a
B
a 6cm
50°
C 7cm A
2. Find c
A
4cm c
Z\
C 5cm B
3. Find all three angles.
C
8cm 6cm
A 11cm B

4. Find all three angles.

7.6cm 4.8cm

S

Cc

B 9.2cm
5. Given that B = 132°, a=5cm and c=5.9cm. Find b
6. Given that € = 33°12’, b=4.5cm and a=6cm. Find c
7. Given that A = 142,2°, b=42cm and c=22cm. Find a

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

8. Given that a=4.5cm, b=3.3cm and c=2.2cm. Find all three angles.
9. Given that a=5.8cm, b=4.1cm, and c=7.2cm, find all three angles of the triangle.
10. Given that a=6.2cm, b=3.5cm, and c=5.9cm, find all three angles of the triangle.

Example 5

13cm W

(a) Find the length of WZ

(b) Find the length of XY

(c) Find angle ZWxX

(d) Find the shortest distance from Z to XW.

(a) Find the length of WZ
We will use triangle WXZ, we two sides and an angle and we are looking for the side
opposite the given angle so we use the cosine rule.

(WZ)? = 10? + 13? — 2(10)(13) cos 40°
= 100 + 169 — 260 cos 40°
= 269 — 260 cos 40°

(WZ)? = 69.8284448 (use a calculator in degrees mode)
V(WZ)2 = V69.8284448 (use calculator)
WZ=8.36cm

(b) Find the length of XY
We will use triangle XYZ, we have two angles and a side so we will use the sine rule to
find XY.

x 13cm Ww
We first find the angle YZX (angle in blue) which is opposite XY.

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

YZX = 180° — (100° + 30°) (angles in a tringle add up to 180°)
= 180° — 130°
= 50°

Now we find XY using the sine rule.
XY _ 10

sin50° sin 100°

xY 10

sin50° x amso > umio0o * sin 50° (multiply both sides by sin 50° to isolate XY)
xY = es (use a calculator in degrees mode)
XY =7.78cm (to 3 significant figures)

(c) Find angle ZW xX
We are going to use triangle WXzZ, to find angle W we can use either the cosine rule or
the sine rule. We are going to use the sine rule.

sinW _ sin 40°

10 8.36

sinW _ sin40°

10 x to aaa * 10 (multiply both sides by 10 to isolate sin W”)
—~ i Oo
sinW = — (use a calculator in degrees mode)
sin W = 50.3° (to one decimal place)

(d) Find the shortest distance from Z to XW.

The shortest distance from Z to XW is perpendicular to XW. To find h (the shortest
distance from Z to XW) we use trig ratios. In the right-angled triangle h is the opposite,
and 10cm is the hypotenuse. So, we use Sine.

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

: _ opposite
sin = hypotenuse
sin 40° =~
10
10 x sin40° = = x 10 (multiply both sides by 10 to isolate h)
10 sin40° =h (use a calculator in degrees mode)
6.43cm=h
~h=6.43cm (to 3 significant figures)
Example 6
X 2x
GD Z
2cm

Y
In the diagram, XYZ is a triangle in which YX = 2 cm, XZ = 2x cm, YZ = 4x cm and YXZ =
120°.
(i) Form an equation in x and show that it reduces to 3x? -x-1=0.
(ii) Hence find the value of x to 3 significant figures.
(iii) Find YZ.

(i) Form an equation in x and show that it reduces to 3x2 --x-1=0.
We use the cosine rule to create the equation.

(4x)? = 2? + (2x)? —2 x 2x x 2 x cos 120°

16x? = 4+ 4x? — 8x x (-0.5) (cos 120° = —0.5)

16x? =44 4x? + 4x [-8x x (—0.5) = 4x]
16x? — 4x? -4x-4=0 (collect everything on the left side)
12

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Online Lessons: 0774 031 978

12x? —4x -—4=0 (16x? — 4x? = 12x?)
4(3x*-x-1)=0 (factor out 4)
4(8x2-x-1) _ 0

4 4
3x2 -x-1=0

(ii) Hence find the value of x

To find x we solve 3x2 — x —1 =0 using the quadratic equation.
a=3,b=-1,c=-1

xo —b+vb2-4ac

2a

—(-D+,}(-*-4«3x(-1)

14V73 1-vi3
ins or

= 0.768cm or —0.434cm

We discard the value —0.434cm because length cannot be a negative.
“x = 0.768cm

(iii) Find YZ.

(divide both sides by 4)

Exam Cheat Code: Maths 4

YZ =4x

= 4 x 0.768

= 3.072cm
Exercise Ie (30 marks] Answers
1.

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

(i) Find AD. [3]
(ii) Find AB. [3]
(iii) Find the shortest distance from B to AC. [2]
(iv) Find the area of triangle ACD. [3]
2.

A

v112
x
60°
X B 2x C

In the diagram, ABC is a triangle in which AB = xcm, BC=2x cm, AC= V112 cm and
ABX = 60°

(i) Form an equation in x and show that it reduces to 7x?-112=0 [3]

(ii) Hence find the value of x. —- [3]

(iii) Find BC. [1]

3. AX is parallel to BC, XAC = 50°
A

50°

B > Cc

(i) Find AB. [3]
(ii) Find the shortest distance from A to BC. [3]

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Online Lessons: 0774 031 978

Exam Cheat Code: Maths 4

(i) Find AD. [3]
(ii) Find ABC to 1 decimal place. [3]

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

Ciswew maths book 4

Exercise 1a back to [Trigonometry]
1. 9.89cm F 2 11.6 |[z 1.ssin 0")

sin 50° sin 64°. sin 64°
2. 77.2° [== mi [sin P = sen [sin P = 0.9751..][P = sin-1(0.9751...)]
5 ° 4 ° o a 10 10sin 62°
3. 8.83cm [B 180 (62° + 30°) = 88 \ sce oll sin 88° ]
4. 55.3° [=* a | [sinc an | [sin C = 0.8217..][C = sin-1(0.8217 ...)]
b 15 15sin60°
5. 20.2cm [aoe md [> sin 40° |
p 0 02¢! 4 0498 0499 q 12 12 sin 45°36"
6. 8.91cm [P 180 (45°36! + 30°12") = 74°12 ] Freer anal | sin 74912"
iG ° O4 0’ — ono n _ 8 _ 8sin 53°
7. 6.49cm [@ = 180° — (53° + 47°) = 80°] [= => | [n — |
8. 64.9° [= mnt | [sin Y vein | [sinY = 0.9057... ][Y = sin-1(0.9057...)]
9. 38° [=* anew | [sin y=venere | [sinY = 0.6151..][Y = sin-1(0.6151...)]
in K in 45° . 24sin 45° . —
10. 58.1° [= ae | [sink = | [sin K = 0.8485..][Y = sin-1(0.8485....)]
Exercise 1b [Trigonometry]

1.5.57cm
[a2 = 6? + 7? — 2(6)(7) cos 50°][a? = 36 + 49 — 84cos50°][a? = 85 — 84cos 50°][a = V85 — 84 cos 50°
2. 2.59cm

[c? = 47 + 5? — 2(4)(5) cos 31°][c? = 16 + 25 — 40.cos 31°][c? = 41 — 40 cos31°][c = V41 — 40 cos 31°]

3. A = 32.29; B = 45.29; € = 102.6°

874112-6? 644121-36 _ 149 1 (149
[cosa = = = | [4 = cos “( |

2x8x11 176 176 176
674+112-87 _ 364121-64 93 -1/ 93
cosB = = = B=cos
2x6x11 132 132 132

[C = 180° — (45.2° + 32.2°)]

4. A = 93°; B = 31.49; € = 55.6°

7.674+14.87-9.2? _ 57.76+23.04-84.64 3.84 -1 (-3.84
cosA = = = A=cos
2X7.6X4.8 72.96 72.96 72.96
7.62+9.27-4.82 _ 57.76+84.64—23.04 119.36 —1 (119.36
cosB = = = B=co!
2X7.6X9.2 139.84 139.84. 139.84,

[c = 180° — (314° + 93°)]

5. 9.96cm

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4 [b? = 5? + 5.9? — 2(5)(5.9) cos 132°] [b? = 25 + 34.81 — 59 cos 132°][b? = 59.81 — 59 cos 132°] [b =

V59.81 — 59 cos 132°|

6. 3.33cm

[c? = 6? + 4.5% — 2(6)(4.5) cos 33°12'][c? = 36 + 20.25 — 54-cos 33°12'][c? = 56.25 — 54.cos 33°12'][c =
56.25 — 54:cos 33°12" |

7. 60.9cm

[a2 = 42? + 22? — 2(42)(22) cos 142.2°][a? = 1764 + 484 — 1848 cos 142.2°][a? = 2248 — 1848cos 142.2°][a =
2248 — 1848 cos 142.2°]

8. A = 108.1°; B = 44.2°; € = 27.7°

3.3742.27-4.5? 10.89+4.84—20.25 _ —4.52 —4 (452
cosA cos
2x3B.3X2.2 14.52 14.52 14.52
4.574+2.22-3.32 20.254+4.84-10.89 14.2 1 (14.2
cosB = = = B= cos
2x4.5X2.2 198 19.8 19.8

[C = 180° — (108.1° + 44.2°)]

9. A = 53.6°; B = 34.7°; € = 91.7°

4.1°+7.27-5.8? _ 16.81+51.84-33.64 35.01 —1 (35.01

cosA A=cos
2X4.1X7.2 59.04 59.04. 59.04
7.274+5.87-4.12 51.844+33.64-16.81 _ 68.67 —1 (68.67

cosB = = = B=co:
2X7.2X5.8 83.52 83.52 83.52

[C = 180° — (53.7° + 34.7°)]

10. A = 78°; B = 33.59; € = 68.5°

3.574+5.97-6.27 _ 12.254+34.81-38.44 __ 8.62 —1 (8.62
cosA = = = A=cos
2x3.5X5.9 41.3 41.3 41.3
6.274+5,97-3.52 38.44434.81-12.25 61 -1f 61
cosB = = = B=cos
2X6.2X5.9 73.16 73.16 73.16,

[C = 180° — (78° + 33.5°)]
Exercise 1¢ [Trigonometry]

1. (i) 20 cm (cosine rule)

[AD? = 12? + 152 — 2(12)(15) cos 95°][AD? = 144 + 225 — 360 cos 95°] [AD? = 369 — 360cos95°][AD =
369 — 360 cos 95°|

(ii) 8.29cm (sine rule)

B j ABs 12 __ 12sin 38°
[B 180° — (25° + 38°)] [Se sin all |
(iii) 3.5cm [sin 25° =] [x = 8.29 sin 25°]
(iv) 89.7cm? [area = +x 12x 15 x sin95? = 90 sin 95°|

2. (i) [(vii2)" = x? + (2x)? — 2(x)(2x) cos(180° — 60°)| [112 x? + 4x? — 4x? x ( *)| [112 =x? 4
4x? + 2x?][112 = 7x?][0 = 7x? — 112]

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Online Lessons: 0774 031 978 Exam Cheat Code: Maths 4

(ii) 4cm [7x? = 112] [x? | |: ~ Vv16 4] is discarded length cannot be negative]

(iii) 8cm. [2x =2x4=8]

3. (i) 4.74cm (cosine rule)

[AB = 52 + 6? — 2(5)(6) cos 50°[AB? = 25 + 36 — 60cos 50°][AB® = 61 — 60 cos 50°][AB = V61 — 60 cos 50°]
(ii) 3.83cm. [sin 50° =;| [x = 5sin 50°]

4. (i) 11.6cm (sine rule)

[ABC 180° — (47° 4 40°] [ AD _ 18 ][40 = S22")

sin 40° sin 93° sin 939
(ii) 82.8° (cosine rule)

5 1274157187 1444225-324 45 1 (45
[cosaBC | [4 cos"? ( )|
2x15x12 360 360. 360.

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