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