Transcript x - Mr-Kuijpers-Math

Solving Problems Modelled by Triangles
US5251
PYTHAGORAS
Can only occur in a right angled triangle
hypotenuse
h
a
Pythagoras Theorem states:
h2 = a2 + b2
b
right angle
e.g.
x
7.65 m
9.4 cm
y
11.3 m
square
root
undoes
squaring
x2 = 7.652 + 11.32
x2 = 186.2125
x = √186.2125
x = 13.65 m (2 d.p.)
8.6 cm
smaller sides
should always
be smaller than
the hypotenuse
y2 + 8.6
9.422==9.4
y2 2+ 8.62
- 8.62 y2 = 9.42-–8.6
8.62 2
y2 = 14.4
y = √14.4
y = 3.79 cm (2 d.p.)
TRIGONOMETRY (SIN, COS & TAN)
- Label the triangle as follows, according to the angle being used.
Hypotenuse (H)
Opposite (O)
Always make
sure your
calculator is set
to degrees!!
A
Adjacent (A)
means divide
1. Calculating Sides
e.g.
7.65 m
x
H
O
29°
O
S
H
to remember the trig ratios use
SOH CAH TOA
and the triangles
x = sin29 x 7.65
x = 3.71 m (2 d.p.)
O
S
A
H
C
O
H
T
A
means multiply
O
50°
6.5 cm A
h
O
T
A
h = tan50 x 6.5
h = 7.75 cm (2 d.p.)
e.g.
H
d
455 m
O
32°
d = 455 ÷ sin32
d = 858.62 m (2 d.p.)
O
S
H
2. Calculating Angles
-Same method as when calculating sides, except we use inverse trig ratios.
e.g.
23.4 mm
H
16.1 mm
O
A
O
S
H
B
sin-1
undoes
sin
2.15 m
A
sinA = 16.1 ÷ 23.4
A = sin-1(16.1 ÷ 23.4)
A = 43.5° (1 d.p.)
Don’t forget brackets, and fractions can also be used
H
cosB = 2.15 ÷ 4.07
A
C
4.07 m
B = cos-1(2.15 ÷ 4.07)
H
B = 58.1° (1 d.p.)
TRIGONOMETRY APPLICATIONS
e.g. A ladder 4.7 m long is
leaning against a wall. The
angle between the wall and
ladder is 27°.
27°
A
C
32 m
A
A
48 m
A
C
H
Ladder (4.7 m)
Draw a diagram and find the
height the ladder extends up
the wall.
H
Wall
(x)
A
H
cosA = 32 ÷ 48
A = cos-1(32 ÷ 48)
A = 48.2° (1 d.p.)
H
x = cos27 x 4.7
x = 4.19 m (2 d.p.)
e.g. A vertical mast is held by a
48 m long wire. The wire is
attached to a point 32 m up
the mast.
Draw a diagram and find the
angle the wire makes with
the mast.
NON-RIGHT ANGLED TRIANGLES
1. Naming Non-right Angled Triangles
- Capital letters are used to represent angles
- Lower case letters are used to represent sides
e.g. Label the following triangle
A
c
B
b
C
a
The side opposite the angle is
given the same letter as the
angle but in lower case.
2. Sine Rule
a = b = c .
SinA SinB
SinC
a) Calculating Sides
e.g. Calculate the length of side p
52° A
6m
b
B 46°
p a
Only 2 parts of the rule are
needed to calculate the answer
p = 6 .
Sin52
Sin46
× Sin52
p = 6 × Sin52
× Sin52
Sin46
p = 6.57 m (2 d.p.)
To calculate you must have the angle opposite
the unknown side.
Re-label the triangle to help substitute info into the formula
b) Calculating Angles
For the statement: 1 = 3 is the reciprocal true?
2 6
Yes as 2 = 6
1 3
Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is
on top and therefore easier to calculate.
You must calculate Sin51
before dividing by 6
(cannot use fractions)
SinA
a == SinB
b == SinC
c .
SinA
a
SinB
b
cSinC
e.g. Calculate angle θ
Sinθ = Sin51
7
6
θ A
6m
b
B 51°
7m a
Sinθ
× 7 = Sin51
× 7× 7
6
θ
θ
To calculate you must have the side
opposite the unknown angle
= sin-1( Sin51 × 7)
6
= 65.0° (1 d.p.)
Re-label the triangle to help substitute info
into the formula
Sine Rule Applications
e.g. A conveyor belt 22 m in length drops sand onto a cone-shaped heap. The
sides of the cone measure 7 m and the cone’s sides make an angle of 32° with the
ground. Calculate the angle that the belt makes with the ground (θ), and the
diameter of the cone’s base (x).
SinA = SinB = SinC
a
b
c
b 7m
32°
116° A
Conveyor belt : 22 m
b
7ma
B 32°
148° B
Aθ
xa
a = b = c .
SinA SinB
SinC
x
= 7 .
Sin116
Sin32
× Sin116x = × Sin116
7 × Sin116
Sin32
x = 11.87 m (2 d.p.)
Sinθ = Sin148
7
22
Sinθ
× 7 = Sin148
×7 ×7
22
θ
θ
= sin-1( Sin148 × 7)
22
= 9.7° (1 d.p.)
3. Cosine Rule
-Used to calculate the third side when two sides and the angle between them
(included angle) are known.
a2 = b2 + c2 – 2bcCosA
a) Calculating Sides
e.g. Calculate the length of side x
13 m b
x
a
A 37°
11 m c
x2 = 132 + 112 – 2×13×11×Cos37
x2 = 61.59
x = √61.59
x = 7.85 m (2 d.p.)
Remember to take square
root of whole, not rounded
answer
Re-label the triangle to help substitute info into
the formula
b) Calculating Angles
- Need to rearrange the formula for calculating sides
CosA = b2 + c2 – a2
2bc
e.g. Calculate the size of the largest angle
P
CosR = 132 + 172 – 242
2×13×17
24 m a
Q
13 m
b
A
Watch you follow the
BEDMAS laws!
17 m c
R
Re-label the triangle to help substitute info into
the formula
CosR = -0.267
R = cos-1(-0.267)
R = 105.5° (1 d.p.)
Remember to use whole
number when taking inverse
Cosine Rule Applications
e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to
the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °.
Calculate the distance from the tee to the hole (x) and the angle (θ) at which the
golfer hit the ball away from the correct direction.
b x a
Hole
Tee
Aθ
130 m
b a
60°A
245 m
c c
Ball
a2 = b2 + c2 – 2bcCosA
x2 = 1302 + 2452 – 2×130×245×Cos60
x2 = 45075
x = √45075
x = 212.31 m (2 d.p.)
CosA = b2 + c2 – a2
2bc
Cosθ = 212.312 + 2452 – 1302
2×212.31×245
Cosθ = 0.848
θ = cos-1(0.848)
θ = 32.0° (1 d.p.)
Remember to use whole number
from previous question!
3D FIGURES
- Pythagoras and Trigonometry can be used in 3D shapes
e.g. Calculate the length of sides x and w and the angles CHE and GCH
A
B
G
F
w
O
7m
H
O
T
A
x2 = 52 + 62
x = √52 + 62
x = √61
x = 7.8 m (1 d.p.)
D
w2 = 72 + 7.82
w = √72 + 7.82
w = √110
w = 10.5 m (1 d.p.)
C
Ax
6m A
E
5m
O
tanCHE = 5 ÷ 6
CHE = tan-1(5 ÷ 6)
CHE = 39.8° (1 d.p.)
O
T
A
Make sure you
use whole
answer for x in
calculation
tanCHE = 7 ÷ 7.8
CHE = tan-1(7 ÷ 7.8)
CHE = 41.9° (1 d.p.)
4. Area of a triangle
- can be found using trig when two sides and the angle between the sides (included
angle) are known
Area = ½abSinC
e.g. Calculate the following area
C
8m
a
9m b
39°
52°
Area = ½×8×9×Sin39
Area = 22.7 m2 (1 d.p.)
89°
Re-label the triangle to help substitute info into
the formula
Calculate size of missing angle using geometry
(angles in triangle add to 180°)
Area Applications
e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to
the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °.
Calculate the area contained in between the tee, hole and ball.
Hole
Tee
Area = ½abSinC
130 m
a
C
60°
Ball
245 m
b
Area = ½×130×245×Sin60
Area = 13791.5 m2 (1 d.p.)