TRIGONOMETRY

Subject:

MATHEMATICS

Term:

First Term

Week:

Week 6

Class:

JSS 3 / BASIC 9

Previous lesson: Pupils have previous knowledge of

EQUATIONS WITH FRACTIONS

that was taught in their previous lesson

Topic:

TRIGONOMETRY

Behavioural Objectives:

At the end of the lesson, learners will be able to

• solve questions on right angle triangle (Revision)
• Explain the relationship between right angle triangles and trigonometric ratios.
• Solve questions on Sine, cosine and tangent of acute angles using tables.
• Applications of trigonometric ratios to finding distances and lengths.

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum
• Workbooks

CONTENT:

WEEK 6

TOPIC:     TRIGONOMETRY

CONTENT:

• Right angle triangle (Revision)
• Relationship between right angle triangles and trigonometric ratios.
• Sine, cosine and tangent of acute angles using tables.
• Applications of trigonometric ratios to finding distances and lengths.

Right Angled Triangle

A right-angled triangle has one of its angles equal to 90^O. That is, if a triangle has one of its three angles equal to 90^O then it is a right-angled triangle. A little square at the corner (angle) signifies that the angle is 90^O.

Right Angled Triangle, or a right-angled triangle, is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right angled triangle is the basis for trigonometry.

A right angled triangle has two other angles apart from the right angle. These are acute angles and are less than 90 degrees. The third angle is always 90 degrees and is called the hypotenuse. It is the longest side in the triangle and is opposite to the right angle.

The other two sides are called the legs of the triangle. They are shorter than the hypotenuse and are adjacent to the right angle.

Right Angled Triangle and the Trigonometric Ratios

A trigonometric ratio is a ratio of the lengths of any two sides of a right – angled triangle.The longest side is called the Hypotenuse, the side facing the required angle is called the opposite and the last side is called the adjacent.

A trigonometric ratio is a way of measuring the size of angles. The ratios are based on the lengths of the sides of a right angled triangle. The most common trigonometric ratios are sine, cosine and tangent. These ratios are usually represented by the symbols sin, cos and tan. The sine ratio is defined as the ratio of the length of the side opposite to an angle to the length of the hypotenuse. The cosine ratio is defined as the ratio of the length of the side adjacent to an angle to the length of the hypotenuse. The tangent ratio is defined as the ratio of the length of the side opposite to an angle to the length of the side adjacent to that angle.

These ratios can be used to calculate the sizes of angles in a right angled triangle. The ratios can also be used to solve problems involving triangles that are not right angled. Trigonometric ratios are used in many areas of mathematics, science and engineering. They are particularly useful in fields such as astronomy, physics and calculus.

The three trigonometric ratios are Sine (sin), Cosine (cos) and Tangent (tan).

The figure below show triangle XYZ in various positions

Where H, A, O represent the Hypotenuse, Adjacent and Opposite respectively.

The sides of the triangles are as follows.

XZ, the hypotenuse, XY, the side opposite to and YZ, the side adjacent to

Using these abbreviations hyp, opp andadj, the Trigonometric ratios are given as:

Sin A =                Cos A =          Tan A =

To obtain the value of any trigonometric ratio, the four-figure table is used.

N.B.: Teacher should teach students how to use the four-figure table to find the values of trigonometric ratios.

Application of Trigonometric Ratios to Finding Distances and Lengths.

Trigonometric ratios can be used to solve real life problems such as angles of elevation and depressions and bearings and other real life situations that result in right-angled triangles.

Use of Sine

Sine and cosine of angles are used to find the lengths of unknown side in triangles. The table below gives the sine of some chosen angles.

The values in the table are given to 4 significant figures.

Example

1. Calculate the value of x in the figure below.

20cm

55⁰

Solution

In the figure above, the hypotenuse (hyp) is given and x is opposite (opp) the given angle. Thus, we can only use:

Sin A =

Sin 55⁰ =

Thus, = 20 x sin 55⁰

= 20 x 0.8192

= 16.384 cm

= 16 cm to 2 s.f.

2. Use tables to find the angles:
3. whose sine is =  ,
4. whose sin is 0.649 2,

Solution

1. Let the angle be A, then sin A = .

Express  as a decimal fraction correct to 4 d.p.

. = 0.285 7

Sin A = 0.285 7

Looking within the sine table entries,

0.2857 is opposite 16⁰ and under 0.6⁰.

Thus, A = 16.6⁰.

1. Let the angle be x, then sin x = 0.649 2.

Looking within the sin table entries, the nearest value to 0.649 2 is 0.648 1. 0.648 1 is opposite 40⁰ and under 0.4⁰.

The difference between 6 492 and 6 481 is 11. Look for 11 in the difference column along the row of 40⁰. 11 is under 8.

Thus, x =   40.40⁰ + 0.08⁰

X = 40.48⁰

Class Activity:

Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.

1. 3.

45⁰

10 cm         x                      y                6 km           Z      3 km

30⁰                                                                 50⁰

Use of Cosine

Example

1. Calculate the value of y in the figure below.

17 cm

50⁰

Solution            

The hypotenuse (hyp) is 17 cm and y is adjacent (adj) to the given angle. Thus use the cosine of the given angle.

Cos A =

Cos 50⁰ =

Thus,  = 17 x cos 50⁰

= 17 x 0.642 8

= 10.93 cm

= 11 cm to 2 s.f

2. Use tables to find the angles:
3. whose cosine is 0.4478,
4. whose cosine is 0.568 2,

Solution

1. Let the angle be B, then cos B = 0.4478.

Looking within the cosine table entries,

0.447 8 is opposite 63⁰ and under 0.4.

Thus, B = 63.4⁰.

1. Let the angle be y, then cos y = 0.568 2.

Looking within the cosine table entries, the nearest value to 0.568 2 is 0.567 8. 0.567 8 is opposite 55⁰ and under 0.4⁰.

The difference between 5 682 and 5 678 is 4.

Look for 4 in the difference column along the row of 55⁰.4 is under 3. As the angles increase, their cosines decrease, therefore subtract the difference.

Thus, y = 55.40⁰ – 0.03⁰

Y = 55.37⁰

Class Activity:

Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.

1. 3.

22mm              x            50⁰

40⁰             12 km            25cm

                                                                Y                                  60⁰

Use of Tangent

Tangents of some chosen angles

Example

1. The angle of elevation of the top of a building is 25⁰ from a point 70 m away on a level ground. Calculate the height of the building.

Solution

The angle of elevation from a point 70 m away on a level ground to the top of a building is 25⁰. The height of the building can be found using the following equation:

Height = Distance * Tan (Angle)

Therefore, the height of the building is:

Height = 70 * Tan (25)

Height = 70 * 0.466

Height = 32.62 m

HK represents the height of the building; AK is on level ground.

H

A         25°                K

70 m

Tan A =

Tan 25⁰ =

Let HK be x cm. KA = 70 cm and, from table above, tan 25⁰ = 0.4663.

Hence,   = 0.4663

X = 0.4663 x70=4.663 x 7

= 32.641

∴The height of the building is 33 m to 2 s.f.

2. Use tables to find the tangents of angles
3. 32⁰ 59.6⁰

Solution

1. Looking within the table entries, the number opposite 32⁰and under 0.0 is 0.6249.

Thus, tan 32⁰ = 0.6249.

1. The number opposite 59⁰ and under 0.6⁰is 1.704. Thus, tan 59.6⁰ is 1.704. Thus, tan 59.6⁰ = 1.704.

Class Activity:

Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.

1. 3.

50⁰

x         12cm      4km              Y   20km

15⁰

5cm                          7km

                                                                                                        30⁰

NOTE: Educators are to solve more examples for students.

Assignment:

1. A village is 8 km on a bearing of 040⁰ from a point O. Calculate how far the village is north of O.
2. Use tables to find the angles whose tangents are:
3. 9556 b.
4. Calculate the length of the hypotenuse of the triangle below.

8 cm

43⁰

4. A car travels 120 m along a straight road which is inclined at 8⁰ to the horizontal. Calculate the vertical distance through which the car rises.

120

8⁰                h

5. A cone is 6 cm high and its vertical angle is 54⁰. Calculate the radius of its base.

Practice Questions:

1. Calculate the lengths a, b, c, d, in the diagrams below. All lengths being in cm.

i                                                          ii 

5         b                               c                         d

37⁰                         40.2⁰

2. Calculate the angles ∝, β in the triangles below.

i                                                                     ii

0.9 m              3m                                     7 m            5 m

∝                                                             β

3. Use table to find the value of each triangles below.

7

35⁰

65⁰

4. Use 4 figure table to find the Sine, Cosine and Tangent of the following:
5. 35⁰ 23.1⁰    c. 19.5⁰
6. Use 4 figure table to find the angles whose Sine, Cosine and Tangent are as follows:
7. 9325 b. 0.8847    c.
8. A cone is 8 cm high and its vertical angle is 62°. Find the diameter of its base.
9. An isosceles triangle has a vertical angle of 116°, and its base is 8 cm long. Calculate its height.
10. Find the angle of elevation of the top of a flagpole 31.9 m high from a point 55 m away on level ground.

PRESENTATION:

Step 1:

The subject teacher revises the previous topic

Step 2:

He or she introduces the new topic

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

CONCLUSION:

The subject goes round to mark the pupil’s notes. He does the necessary corrections