us5251
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Solving Problems Modelled by Triangles. US5251. PYTHAGORAS. Can only occur in a right angled triangle. hypotenuse. Pythagoras Theorem states:. h. a. h 2 = a 2 + b 2. b. right angle. e.g. x. 9.4 cm. 7.65 m. y. 11.3 m. 8.6 cm. x 2 = 7.65 2 + 11.3 2. y 2 + 8.6 2 = 9.4 2. - PowerPoint PPT PresentationTRANSCRIPT
Solving Problems Modelled by Triangles
PYTHAGORASPYTHAGORASCan only occur in a right angled triangle
Pythagoras Theorem states:
hypotenuse
right anglee.g.
square root undoes squaring
smaller sides should always be smaller than the hypotenuse
h2 = a2 + b2
ha
b
x7.65 m
11.3 m
9.4 cm y
8.6 cm
x2 = 7.652 + 11.32x2 = 186.2125 x = √186.2125 x = 13.65 m (2 d.p.)
9.42 = y2 + 8.62y2 + 8.62 = 9.42
- 8.62- 8.62y2 = 9.42 – 8.62
y2 = 14.4y = √14.4
y = 3.79 cm (2 d.p.)
TRIGONOMETRY (SIN, COS & TAN)TRIGONOMETRY (SIN, COS & TAN)- Label the triangle as follows, according to the angle being used.
A
Hypotenuse (H)Opposite (O)
Adjacent (A)
to remember the trig ratios use SOH CAH TOA
and the triangles
S
O
H C
A
H T
O
Ameans divide
means multiply 1. Calculating Sides
29°
e.g.
x7.65 m H
O
S
O
H
x = sin29 x 7.65x = 3.71 m (2 d.p.)
50°6.5
cm
hO
A
T
O
A
h = tan50 x 6.5h = 7.75 cm (2 d.p.)
Always make sure your calculator is set to degrees!!
e.g.
d455 m
32°
HO S
O
H
d = 455 ÷ sin32 d = 858.62 m (2 d.p.)
2. Calculating Angles
-Same method as when calculating sides, except we use inverse trig ratios.
A
16.1 mm
23.4 mme.g.
OH
S
O
H
sinA = 16.1 ÷ 23.4
sin-1 undoes sin
A = sin-1(16.1 ÷ 23.4)A = 43.5° (1 d.p.)
Don’t forget brackets, and fractions can also be used
B2.15 m
4.07 mH
A
C
A
H
cosB = 2.15 ÷ 4.07B = cos-1(2.15 ÷ 4.07)B = 58.1° (1 d.p.)
TRIGONOMETRY APPLICATIONSTRIGONOMETRY APPLICATIONSe.g. A ladder 4.7 m long is leaning against a wall. The angle between the wall and ladder is 27°. Draw a diagram and find the height the ladder extends up the wall. 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.
Wall (x)
Ladder (4.7 m)
27°
H
AC
A
H
x = cos27 x 4.7x = 4.19 m (2 d.p.)
48 m32
m
AH
A
C
A
H
cosA = 32 ÷ 48A = cos-1(32 ÷ 48)A = 48.2° (1 d.p.)
NON-RIGHT ANGLED TRIANGLESNON-RIGHT ANGLED TRIANGLES1. 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
B
C
The side opposite the angle is given the same letter as the angle but in lower case.
b
cA
2. Sine Rule
a = b = c .SinA SinB SinC
a) Calculating Sides
e.g. Calculate the length of side p
p
6 m
52°46°
To calculate you must have the angle opposite the unknown side.
Only 2 parts of the rule are needed to calculate the answer p = 6 .Sin52 Sin46 × Sin52 × Sin52 p = 6 × Sin52
Sin46
p = 6.57 m (2 d.p.)
Re-label the triangle to help substitute info into the formula
AB
ab
b) Calculating AnglesFor 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.
a = b = c .SinA SinB SinC
SinA = SinB = SinC a b c
e.g. Calculate angle θ
7 m
6 m
θ51°
Sinθ = Sin51 7 6
To calculate you must have the side opposite the unknown angle
× 7 × 7Sinθ = Sin51 × 7 6
θ = sin-1( Sin51 × 7) 6θ = 65.0° (1 d.p.)
You must calculate Sin51 before dividing by 6 (cannot use fractions)
Re-label the triangle to help substitute info into the formula
AB
ab
Sine Rule Applicationse.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).
Conveyor belt : 22 m
θ
7 m 7 m32° 32°
x
148° A
a b
B
Sinθ = Sin148 7 22
× 7 × 7Sinθ = Sin148 × 7 22
θ = sin-1( Sin148 × 7) 22θ = 9.7° (1 d.p.)
SinA = SinB = SinC a b c
a = b = c .SinA SinB SinC
116° A
a
bB
x = 7 .Sin116 Sin32
× Sin116
× Sin116
x = 7 × Sin116 Sin32 x = 11.87 m (2 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
x37°
13 m
11 m
Re-label the triangle to help substitute info into the formula
a
A
b
c
x2 = 132 + 112 – 2×13×11×Cos37x2 = 61.59
x = √61.59
x = 7.85 m (2 d.p.)
Remember to take square root of whole, not rounded answer
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
R
Q
17 m
24 ma
Abc
Re-label the triangle to help substitute info into the formula
CosR = 132 + 172 – 242
2×13×17
Watch you follow the BEDMAS laws!
CosR = -0.267
Remember to use whole number when taking inverse
R = cos-1(-0.267)
R = 105.5° (1 d.p.)
13 m
Cosine Rule Applicationse.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.
Hole
245 m
Tee
Ball
130 m
x
θ
60°
a
Ab c
a2 = b2 + c2 – 2bcCosA
x2 = 1302 + 2452 – 2×130×245×Cos60x2 = 45075
x = √45075
x = 212.31 m (2 d.p.)
CosA = b2 + c2 – a2
2bc
Cosθ = 212.312 + 2452 – 1302
2×212.31×245Cosθ = 0.848
θ = cos-1(0.848)
θ = 32.0° (1 d.p.)
A
a
b
c
Remember to use whole number from previous question!
3D FIGURES3D 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
x
w
6 m
7 m
H
G F
E
D C
BA
5 m
x2 = 52 + 62
x = √52 + 62
x = √61x = 7.8 m (1 d.p.)
w2 = 72 + 7.82
w = √72 + 7.82
w = √110w = 10.5 m (1 d.p.)
Make sure you use whole answer for x in calculation
OA
tanCHE = 5 ÷ 6 CHE = tan-1(5 ÷ 6) CHE = 39.8° (1 d.p.)T
O
A
O
A
T
O
A
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
52°
89°
8 m
9 m
Re-label the triangle to help substitute info into the formula
C
a
b
39°
Calculate size of missing angle using geometry (angles in triangle add to 180°)
Area = ½×8×9×Sin39
Area = 22.7 m2 (1 d.p.)
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
245 m
Tee
Ball
130 m 60° a
C
b
Area = ½×130×245×Sin60 Area = 13791.5 m2 (1 d.p.)
Area = ½abSinC