SAMPLE

Applying a progression of steps to

achieve the required answer (φ37)

CCE Lesson

Phi 37

Tricky Trigonometry

Irregular Running

Resource code: 27054550SAMPLE

SAMPLE

SAMPLE

Tricky Trigonometry

In her laboratory, Anna, a budding physicist, is trying to calculate the results from her latest experiments. Her

experiments aim to validate various trigonometric processes, rules and identities.

TIP

Remember that 180 degrees is

equal to π radians.

!

a) tan 5π

6

b) sin π

c) cos 315 °

Q1 Some of her results require her to calculate the values of angles. Without the use of a calculator,

determine the exact value of the following angles. Be sure to show all working including diagrams in

the blank space required. Use the triangles at the bottom of the page to assist you.

45°

45°

1

1

√2

√3

60°

30°

2

1

SAMPLE

Tricky Trigonometry

Q2 In the middle of a square field with a perimeter of 200 m, the angle of elevation from Anna to the top of

a tree in a corner of the field was 47° whilst the angle of elevation to a bird in the tree was half of this

angle. Calculate the height of the tree above the bird. Make sure you draw diagrams and state any

assumptions.

TIP

Give your answer to 2 decimal

places.

!

SAMPLE

Irregular Running

TIP

Draw a diagram to assist you solve

this problem.

!

Q1 Anna started her afternoon run from her home, where she ran 4.3 km north-east to the park. She then

ran 1.5 km on a true bearing of 121° to the local shops where she returned home in a straight line.

Determine the total length of her run and the true bearing on which she returned home from the shops.

Anna isn’t only a keen mathematician/scientist. She also runs every afternoon and she can’t help but think about the

trigonometry of her running paths!

SAMPLE

Irregular Running

Q2 On the next day Anna starts her run at home and sets off in a direction N50°W for 2.0km. She stops for

15 minutes before turning and running N70°E to a position due North of her home. After a half hour

snack, she turns and runs home. If she ran at a speed of 3 m/s, how long was she away from home?

TIP

The formula for speed is speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒. Remember to draw a

diagram and answer in hours!

!

SAMPLE

Applying a progression of steps to

achieve the required answer (φ37)

CCE Lesson

Phi 37

Tricky Trigonometry

Irregular Running

Resource code: 27054550SAMPLE

Please note: any activity that is not completed during class time may be set for homework or

undertaken at a later date.

‘Tricky Trigonometry’ and ‘Irregular Running’

• Activity Description:• This lesson contains two activities that have been designed to allow students to construct

answers for increasing difficult trigonometric problems by applying a progression of steps to

achieve the required answer.

• The first activity, ‘Tricky Trigonometry’, required students to use the unit circle and

exact triangles to calculate trigonometric functions without a calculator and also

involves a complex problem solving question.

• The second activity, ‘Irregular Running’, extends students’ knowledge of

trigonometry with problem solving questions that required them to work in certain

steps in order to reach the answers.

• Purpose of Lesson:• To enhance students’ ability to work in the correct sequence of steps in

order to reach the correct answer and further develop and reinforce

trigonometric knowledge .

• KLAs:• Mathematics B

• CCEs:• Applying a progression of steps to achieve the required answer (Φ37)

• Suggested Time Allocation:• This lesson is designed to take approximately 40 minutes to complete.

• Teaching Notes:• This lesson requires a progression of steps to be performed in order to solve mathematical

problems.

• A pen/pencil is required, as well as a calculator.

• Students should not be able to use their calculators in Question One.

• Angle rules such as the ‘Z rule’ are assumed knowledge throughout the lesson.

• Always encourage students to draw a diagram of problem solving, worded

questions.

• If struggling, ensure students list their known variables and what variables they

need to find out.

• Follow up/ Class Discussion Questions:• Where could you use trigonometry in real life?

• What are some other things in mathematics that need a progression of steps to find the

answer?

Item Description

SAMPLE

In her laboratory, Anna, a budding physicist, is trying to calculate the results from her latest experiments. Her

experiments aim to validate various trigonometric processes, rules and identities.

5𝜋

6x 180

𝜋= 150°, 150° is in the second quadrant. 180° - 150° = 30°

TIP

Remember that 180 degrees is

equal to π radians.

!

Tricky Trigonometry

a) tan 5π

6

b) sin π

c) cos 315 °

45°

45°

1

1

√2

√3

60°

30°

2

1

Q1 Some of her results require her to calculate the values of angles. Without the use of a calculator,

determine the exact value of the following angles. Be sure to show all working including diagrams in

the blank space required. Use the triangles at the bottom of the page to assist you.

x

ySin + ve

Tan + ve

All + ve

Cos + ve

150°

𝜋 ×180

𝜋= 180°

Using knowledge of the unit circle, it is known that sin π will be equal to the y-

coordinate at the 180 degree mark on the plane.

The y-coordinate is 0 (as it is on the x-axis) and has no vertical component.

Therefore, sin π = 0

x

ySin + ve

Tan + ve

All + ve

Cos + ve

180°

315° is in the fourth quadrant. Therefore cos (315)° is positive.

360° - 315 ° = 45°

x

ySin + ve

Tan + ve

All + ve

Cos + ve

30°

45°

From the triangle below, tan (30°) = 1

3

Therefore, tan (5𝜋

6) = −

1

3

From the triangle below, cos (45°) =1

2

Therefore, cos (315°) =1

2

tan(5𝜋

6) is negative (because it’s in the second quadrant).

SAMPLE

200 ÷ 4 = 50 m. Each side is 50 m.

Half of a side: 50 ÷ 2 = 25 m.

Distance from middle to corner = w

w2 = 252 + 252 = 1250

w = 1250= 25 2m

Q2 In the middle of a square field with a perimeter of 200 m, the angle of elevation from Anna to the top of

a tree in a corner of the field was 47° whilst the angle of elevation to a bird in the tree was half of this

angle. Calculate the height of the tree above the bird. Make sure you draw diagrams and state any

assumptions.

TIP

Give your answer to 2 decimal

places.

!

Tricky Trigonometry

50 m

25 m

25 m

25 √2 m

23.5°47°

x m

z m

y m

w m

bird

tan 47° = 𝑧

25 2

z = 25 2 × tan 47° = 37.91 m

The total height of the tree is 37.91 m

tan 23.5° = 𝑦

25 2

y = 25 2 × tan 23.5° = 15.37 m

The height of the tree below the bird is 15.37 m.

x + y = z

x = z – y

x = 37.91 – 15.37 = 22.54 m

The height of the tree above the bird is 22.54 m. It has

been assumed that the distance between Anna’s eyes

and the ground is negligible. In other words, the angle of

elevation is measured from the ground.

SAMPLE

Q1 Anna started her afternoon run from her home, where she ran 4.3 km north-east to the park. She then

ran 1.5 km on a true bearing of 121° to the local shops where she returned home in a straight line.

Determine the total length of her run and the true bearing on which she returned home from the shops.

Irregular Running

N

121°

59°45°

Cosine rule: c2 = a2 + b2 – 2abcos(C)

Let: home ->park = a, park -> shops = b, home ->

shops = c, angle C = 45° + 59° = 104°

Therefore, c2 = 4.32 + 1.52 - 2 × 4.3 × 1.5 × cos(104°) = 23.86

If c2 = 23.86, then c = 23.86 = 4.88km

True bearing = 360° - β° - θ° β° = 59° because of the z rule as shown in the diagram above

True bearing is then = 360° - 59° - 58.76° = 242.2°

So, total length of the run = 4.3 +1.5 + 4.88 = 10.7 km

Bearing of last leg = 242°T

Home

Park

Shops

TIP

Draw a diagram to assist you solve

this problem.

!

θ°

Sine rule: sin(𝐴)

𝑎=

sin(𝐵)

𝑏,

sin(θ)

4.3=

sin(104°)4.88

, θ = 𝑠𝑖𝑛−1sin 104 ×4.3

4.88= 58.76°

Park

θ°

N

β°

Shops

β°

Anna isn’t only a keen mathematician/scientist. She also runs every afternoon and she can’t help but think about the

trigonometry of her running paths!

SAMPLE

TIP

The formula for speed is speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒. Remember to draw a

diagram and answer in hours!

!

Irregular Running

Q2 On the next day Anna starts her run at home and sets off in a direction N50°W for 2.0km. She stops for

15 minutes before turning and running N70°E to a position due North of her home. After a half hour

snack, she turns and runs home. If she ran at a speed of 3 m/s, how long was she away from home?

N

50°

C°

70°

Home

b

c

A°

Sine rule: sin(𝐴)

𝑎=

sin(𝐵)

𝑏,

2

sin(70°)=

b

sin(50°)

Total distance = a + b + c = 2 + 1.63 + 1.84 = 5.47km or 5470m

𝑏 =sin 50° × 2

sin(70°)= 1.630𝑘𝑚

A° = 70° because of the Z rule

Sine rule: sin(𝐴)

𝑎=

sin(𝐵)

𝑏,

2

sin(70°)=

c

sin(60°)

C° = 180° - 70° - 50° = 60 °

𝑐 =sin 60° × 2

sin(70°)= 1.84𝑘𝑚

Total time running = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑠𝑝𝑒𝑒𝑑=

5470

3= 1823 𝑠 ÷ 60 = 30.4 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

Total time from leaving to returning home = 30.4 + 15 + 30 = 75.4 minutes ÷ 60 = 1.26 hours

SAMPLE

Tricky Trigonometry

Question One:

Students were required to find the value of certain trigonometric functions using their knowledge of

both the unit circle and exact triangles. All working was to be shown. A model response is provided

below.

Model Response:

a)5𝜋

6x 180

𝜋= 150°, 150° is in the second quadrant. 180° - 150° = 30°.

From the triangle below, tan (30°) = 1

3

tan(5𝜋

6) is negative (because it’s in the second quadrant).

Therefore, tan (5𝜋

6) = −

1

3

b) 𝜋 ×180

𝜋= 180°

Using knowledge of the unit circle, it is known that sin π will be equal to the y-

coordinate at the 180 degree mark on the plane.

The y-coordinate is 0 (as it is on the x-axis) and has no vertical component. Therefore,

sin π = 0

c) 315° is in the fourth quadrant. Therefore cos (315)° is positive.

360° - 315 ° = 45°

From the triangle below, cos (45°) =1

2

Therefore, cos (315°) =1

2

This teacher’s answer guide is continued on the next page...

Sin + ve

Tan + ve

All + ve

Cos + ve

150°

30°

x

ySin + ve

Tan + ve

All + ve

Cos + ve

180°

x

y

x

y

Sin + ve

Tan + ve

All + ve

Cos + ve

180°

SAMPLE

Tricky Trigonometry

Question Two:

Students were required to set up relations between Anna’s current location in the middle of a square

field and the position of a tree in the corner of the field. This was done using the Pythagorean

Theorem. Then the tangent function was used to find the total height of a tree and the height of where

a bird was. The difference of these two figures was then found to calculate the distance from the bird

to the top of the tree. A model Response is provided below.

Model Response:

200 ÷ 4 = 50 m. Each side is 50 m.

Half of a side: 50 ÷ 2 = 25 m.

Distance from middle to corner = w

w2 = 252 + 252 = 1250

w = 1250= 25 2m

tan 47° = 𝑧

25 2

z = 25 2 × tan 47° = 37.91 m

The total height of the tree is 37.91 m

tan 23.5° = 𝑦

25 2

y = 25 2 × tan 23.5° = 15.37 m

The height of the tree below the bird is 15.37 m.

x + y = z

x = z – y

x = 37.91 – 15.37 = 22.54 m

The height of the tree above the bird is 22.54 m. It has been assumed that the distance between

Anna’s eyes and the ground is negligible. In other words, the angle of elevation is measured from

the ground.

50 m

25 m

25 m

25 √2 m

23.5°47°

x m

z m

y m

w mbirdSA

MPLE

Irregular Running:

Question One:

Students were required to draw the path of Anna’s run and write in all the correct navigation

information such as angles and lengths. This diagram and information was then used in the cosine

function to ascertain the length of the unknown side of the triangular running route. After this was

done the missing angle was found and the true bearing of the final leg of the route was taken. The

three different legs of the running path were added to find the total distance covered. Full working

should have been shown with a progression of steps. A model response is shown below.

Model Response:

Cosine rule: c2 = a2 + b2 – 2abcos(C)

Let: home ->park = a, park -> shops = b, home -> shops = c, angle C = 45° + 59° = 104°

Therefore, c2 = 4.32 + 1.52 - 2 × 4.3 × 1.5 × cos(104°) = 23.86

If c2 = 23.86, then c = 23.86 = 4.88km

Sine rule: 𝑠𝑖𝑛(𝐴)

𝑎=

𝑠𝑖𝑛(𝐵)

𝑏,

𝑠𝑖𝑛(𝜃)

4.3=

𝑠𝑖𝑛(104°)4.88

, 𝜃 = 𝑠𝑖𝑛−1𝑠𝑖𝑛 104 ×4.3

4.88= 58.76°

True bearing = 360° - β° - θ° β° = 59° because of the z rule as shown in the diagram

below

True bearing is then = 360° - 59° - 58.76° = 242.2°

So, total length of the run = 4.3 +1.5 + 4.88 = 10.7 km

Bearing of last leg = 242°T

This teacher’s answer guide is continued on the next page...

N

121°

59°45°

Home

Park

Shopsθ°

Park

θ°

N

β°

Shops

β°

SAMPLE

The Numbers Don’t Lie

Question One:

Students where required to find the length of the sides of a triangular running path in order to

calculate how long it took Anna to run the entire length. This was done by applying the z rule to find a

missing angle and also the sine rule to find missing lengths. Given the formula speed=distance/time,

the time taken to run the route was then found and added on to the time of two breaks. This found the

total time. Full working should have been shown. A model response is shown below.

Model Response:

A° = 70° because of the Z rule

𝑠𝑖𝑛(𝐴)

𝑎=

𝑠𝑖𝑛(𝐵)

𝑏,

2

𝑠𝑖𝑛(70°)=

𝑏

𝑠𝑖𝑛(50°)

𝑏 =𝑠𝑖𝑛 50° × 2

𝑠𝑖𝑛(70°)= 1.630𝑘𝑚

C° = 180° - 70° - 50° = 60 °

Sine rule: 𝑠𝑖𝑛(𝐴)

𝑎=

𝑠𝑖𝑛(𝐵)

𝑏,

2

𝑠𝑖𝑛(70°)=

𝑐

𝑠𝑖𝑛(60°)

𝑐 =𝑠𝑖𝑛 60° × 2

𝑠𝑖𝑛(70°)= 1.84𝑘𝑚

Total distance = a + b + c = 2 + 1.63 + 1.84 = 5.47km or 5470m

Total time running = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑠𝑝𝑒𝑒𝑑=

5470

3= 1823 𝑠 ÷ 60 = 30.4 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

Total time from leaving to returning home = 30.4 + 15 + 30 = 75.4 minutes ÷ 60 = 1.26 hours

This teacher’s answer guide is continued on the next page...

N

50°

C°

70°

Home

b

c

A°

SAMPLE