Trigonometry Part III

Course Description

Level: High school (Grade 10/11)

This section is an extension of Part I & II where we deal with real life applications of trigonometry. It is well known that trigonometry has been part of human life from as early as ancient civilizations, especially in Asia and Europe. From the basics of right angled triangle to the sine graph, trigonometry has wide range of applications, in many fields from engineering to rocket science.

Happy Learning!

Prior Knowledge required:

• Trigonometry Part I
• Trigonometry Part II
• Ability to understand questions based on real-life situations and convert to mathematical form
• Ability to sketch diagrams related to the word problems
• Use of graphing display calculator or graphing software.

Trigonometry Part III topic links

• Unit I – Simple applications
• Unit II – Bearings
• Unit III – Angles of Elevation and Depression
• Unit IV – 3D Trigonometry
• Unit V -Modeling using trigonometric functions

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Unit I – Simple Applications

In this unit, you will see some real life problems that use right triangle trigonometry. First step in solving all these problems is to sketch an appropriate diagram for the problem.

Some of the vocabulary and their related meaning in mathematical terms in these types of problems are as follows:

1. Ground – horizontal side
2. Wall – vertical side
3. Leaning against the wall – hypotenuse
4. Slant roof – hypotenuse

 

Question 1:

A ladder of length 20 feet leans on a vertical wall at an angle of 30⁰ with the ground. Find the distance between the wall and the foot of the ladder.

Solution:

Converting the above problem into a diagram, the vertical wall would be a vertical leg and the ground will be the horizontal leg. The ladder would be the hypotenuse as shown below:

The triangle ABC on the right is clearly a right angled triangle with

AC = 20,\angle{ACB} = 30^0

The distance between the wall and the foot of the ladder is BC.

Connecting all the above information, we can use simple trigonometric ratios to find BC.

­\cos(\angle{ACB})=\dfrac{BC}{AC}

­\cos(30^0)=\dfrac{BC}{20}

­\cos(30^0)\times 20 =BC

­BC= \dfrac{\sqrt3}{2}\times 20 = 10\sqrt{3} feet

Question 2:

A flag pole of height 100 m is supported by a wire tied from its top to a point on the ground 10m away from the foot of the pole. What is the angle that this wire makes with the ground?

Solution:

Clearly, the wire makes the hypotenuse, while the flag post itself is the vertical leg. As before, the distance between the foot of the flag pole and the point where the wire is tied to the ground makes the horizontal leg.

Now, let us draw the diagram for the above question:

The angle to be found is θ and can be found using tan ratio since the opposite side for this angle is the flag post and the adjacent side is the ground distance.

Hence,

­\tan(\theta) = \dfrac{100}{10} = 10

­\theta = \arctan(10) \approx 84.3^0

Question 3:

A café  needs a sunshade for their customers who prefer to sit outside the shop. The width of the seating area is 1.5 meters and the sunshade should be able to cover this width. The height of the front wall of the shop is 3.75 meters. The lower part of the sunshield should be at 2.5 meters from the ground to give enough head space for the customers. What is the angle that the sunshade should be aligned to the front wall?

Solution:

First it is required to identify the right angled triangle from the problem. Since the question is to find the angle that the shade makes with the vertical wall, the triangle made by the sunshade with the wall is to be considered as shown below:

The remaining length of 1.25 m on the vertical wall makes one leg of the triangle while the horizontal distance (1.5 m) makes the other leg. The angle θ is the required angle and it can be found as follows:

­ \tan(\theta) = \dfrac{1.5}{1.25}

­\theta = \tan^{-1} (\dfrac{1.5}{1.25}) \approx 50\degree

Hence the sunshade should be at an angle of 50\degree to the vertical wall.

 

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Unit II – Bearings

Bearings are the type of measurement used to indicate directions and are commonly used for navigation purposes. You might have often noticed a needle or an arrow mark showing the north in all the navigation maps, such as google map and GPS. Have you ever wondered why only north is specifically displayed? Why not east? Or west or south?

This is because the direction is measured as an angle from the north, clockwise as shown below:

Question 1:

A runner runs due east for 15 km and then turns to run towards south for 12 km. Find the bearing of his final position from his starting position.

Solution:

Let us first draw the diagram representing the information in the question:

The required bearing is the angle θ as marked on the diagram. We know that the angle between the north and the east is 90\degree and hence it is enough to find the part of the angle θ that is inside the triangle. Since the triangle is a right angled one, we can find this angle, say α, as below:estion:

\tan (\alpha) = \dfrac{12}{15}\Rightarrow \alpha \approx 38.7\degree­

Therefore, the required bearing is

90 + 38.7 = 128.7\degree

 

Question 2:

Arron walks in the direction of north at a speed of 6km/hr and Ben starts from the same point as Arron, but runs at a speed of 10 km/hr. Find the distance between Arron and Ben after 90 minutes.

Solution:

Note the change in the  units for time in the question. The 90 minutes should be changed to 1.5 hours to suit the unit for speed.

Next, let us draw the necessary diagram as shown below:

Arron’s distance along north after 1.5 hours would be 9 km and Ben’s along west would be 15 km.

The distance to be calculated here is the side AB of the right angled triangle OAB. Using Pythagoras theorem,

AB = \sqrt {15^2 + 9^2} \approx 17.5 km.

 

Question 3:

A boat moves at a bearing of 65\degree for 30 km. Find how far is the boat north of its starting position.

Solution:

In this question, the distance to be found is OB, which gives the boat’s position to the north of its starting point.

\dfrac{OB}{OA} = \cos(65\degree)

Therefore, OB = \cos 65\degree \times OA OB \approx 12.7 km

 

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So far you saw how bearing problems can be solved using basic trigonometric ratios. You should have noticed that in all the above questions the movement were in one of the four directions – east, west, north or south – which created right angled triangle. Therefore, it was possible to use basic ratios. However, if you notice the first example of the boat moving from port X to port Y seen in the video above, you can see that there is no movement along any of the four directions, but with different bearings. For such situations, we use cosine and sine rules. This is explained in the video below:

Note: In any bearing questions, it is necessary to draw diagrams representing the situation in the problem and then convert them to mathematical representation such as triangle XYZ etc. as shown in the above videos

 

Question 1:

An airplane travels from a point A at a bearing of 125⁰ for a distance of 150 km. Then it changes its direction and travels at a bearing of 215⁰ for a distance of 100 km to a point B. Find the distance between point A and point B.

Solution:

Let us first draw the diagram representing the information in the question:

After drawing the diagram, notice that the corresponding angle between the two parallel north lines will give 55⁰ and hence \angle ACB = 90\degree

Therefore, ABC is a right angled triangle and we can use Pythagoras theorem to find x

­x^2 = 100^2 + 150^2 = 32500 \Rightarrow x \approx 180.2 km.

 

Question 2:

A ship is sailing towards east for a distance of 300 km and then changes direction towards south and travels for a distance of 120 km. Find the bearing of the final position of the boat from its starting position.

Solution:

First step is to draw the diagram for the given problem. From the diagram, it is clear that the required bearing is the angle shown in green. To find this angle, it is enough to find the angle AOB in the right triangle (shown as question mark)

\angle {AOB} = \arctan \left(\dfrac{120}{300}\right)\approx 21.8\degree

Therefore, the required bearing is 90\degree + 21.8\degree = 111.8\degree

 

Question 3:

A car drives at 120 km/h for 50 minutes on a bearing of 50° and then 150 km/h for 30 minutes on a bearing of 300°. Find the distance and bearing of the car from its starting point.

Solution:

The diagram for the given problem is shown here. The distances travelled are calculated using the speed and time given in the question.

AB = 100 km and BC = 75 km

Using the concept of alternate angles on the two north lines, we have \angle {ABC} = 70\degree

In order to find the distance of the ship from its starting position (x), we can use Cosine rule on triangle ABC.

x = \sqrt{100^2+75^2-(2\times 100\times75\times \cos(70\degree))}

­­ \Rightarrow x \approx 102 km

­The bearing (θ) of this direction =  50\degree - \angle {CAB}

­\angle {CAB} can be found using Sine rule as shown below:

­\dfrac{\sin{\angle {CAB}}}{75}=\dfrac{\sin{70\degree}}{102}

­­\angle{CAB}\approx 43.7\degree

­Hence \theta = 50 - 43.7 =6.3\degree

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Unit III – Angles of Elevation and Depression

This section deals with how Trigonometry is used in questions where angles are formed when we look up or look down from a given point. For example, when you look up at the top of the street light pole from the ground, you raise your head. Similarly, standing on the top of a building, you need to bend your head down to look at a car on the street. In both these cases, the head is raised or lowered from the horizontal view and hence an angle is formed with the horizontal. This is explained more clearly in the video below:

Note: For questions involving angle of elevation and depression, it is necessary to draw diagrams representing the situation in the problem and then convert them to mathematical representation such as triangle XYZ etc. as shown in the above videos

More examples on the above concept:

Question 1:

From a point on the ground 25 m away from the foot of a flag pole, the angle of elevation to the top of the flag pole is 40°. Find the height of the pole

Question 2:

From a point A on the ground, the angle of elevation of the top of a 15 m tall building is 30°. A flag pole is placed at the top of the building and the angle of elevation of the top of the flag pole from A is 35°. Find the height of the flag pole and the distance of the building from point A. Give your answer in 3 significant figures.

Question 3:

From the top of a tall tower, the angles of depression of the top and bottom of a 10 m tall building are 40° and 60° respectively. Find the height of the tower and the distance between the tower and the building.

Question 4:

Two buildings of equal height are standing opposite to each other on either side of a road 100 m wide. From a point between them on the road, the angle of elevation of the top of the buildings are 75° and 35°. Find the height of the buildings and the distance of the point from the buildings.

The solutions to the above 4 examples can be found in this video

 

 

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Unit IV – 3 D Trigonometry

In this section we will deal with 3 – dimensional problems that can be solved using trigonometry. A typical 3 – D figure is a cube or a cuboid. You can imagine your room to be a cuboid and draw imaginary diagonals to visualize this concept. A visual approach to this 3-dimensional problems is given in the video below:

 

Let us now solve some problems based on the above:

Examples on the above concept:

Question 1:

In the cuboid shown here, find the angle between AG and the base, given that the dimension of the base is 5 cm X 4 cm and the height of the cuboid is 4.5 cm.

Solution:

By joining the vertices A and G, we get the diagonal AG. The angle this line makes with the base is \angle AGE in the right angled triangle AGE. In this triangle, \angle AEG= 90\degree as shown below

EG is the diagonal of the base and its length can be found using Pythagoras theorem. The base dimensions are 5 cm X 4 cm.

Therefore, EG = \sqrt {5^2 +  4^2} \approx 6.4 cm

Now using the triangle AEG,

\tan(\angle AGE) =\dfrac{AE}{EG}= \dfrac{4.5}{6.4} \angle AGE=\tan^{-1} \dfrac{4.5}{6.4} \approx 35.1\degree

 

Question 2:

The cuboid shown below has base 9 cm X 5 cm and height 4 cm. Find \angle HAF.

Solution:

The triangle AHF is formed by the diagonals of three sides of the cuboid. Hence, the lengths of these sides can be found using Pythagoras theorem on the three faces. We have HE = 5cm; HG = 9 cm; AE = 4 cm.

AH =\sqrt {AE^2+HE^2} = \sqrt{4^2+5^2} \approx 6.4

­HF =\sqrt {HG^2+GF^2} = \sqrt{9^2+5^2}\approx 10.3

­AF =\sqrt {AE^2+EF^2} = \sqrt{4^2+9^2} \approx 9.8

To find \angle HAF, cosine rule can be applied:

\angle HAF = \cos^{-1}\left[\dfrac{6.4^2+9.8^2-10.3^2}{2\times6.4\times9.8}\right]\approx 75.7\degree

Question 3:

In the pyramid shown  below, find the height x and \angle {EAC}

 

 

 

 

 

 

 

 

Solution:

To find the height of the pyramid, we consider the right angled triangle ANC (since AN is perpendicular to EC)

­NC = \frac{1}{2} EC = \frac{1}{2}\sqrt{ED^2+DC^2} = \frac{1}{2}\sqrt{9^2+7^2}= 5.7 cm

­AN=\sqrt{AC^2-NC^2} = \sqrt{16^2-5.7^2}\approx 15 cm

To find \angle EAC:

­\angle EAC= 2\angle NAC = 2 \tan^{-1}\dfrac{5.7}{15}\approx 41.6\degree

 

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Unit V – Modeling using trigonometric functions 

Any real life situation which occurs periodically, such as, temperature in a city over a period of a year, the distance a pendulum traces while swinging, the path of a particle when it goes round a circle (eg: Ferris wheel) etc. can be modelled using trigonometric function. Generally, we use sine function to model such situations, but occasionally cosine function will be used as suited.

Following is an example of a periodic model:

The monthly average temperature data of Tokyo city over a year is given below (in ºC):

Jan	Feb	Mar	Apr	May	June	July	Aug	Sep	Oct	Nov	Dec
5	6	9	14	18	21	25	26	23	18	12	8

[Source:https://www.holiday-weather.com/tokyo/averages/]

Plotting the points on a graph, by considering Jan = 1, Feb = 2 etc. on the x-axis and  ºC on the y-axis, we get a graph as shown below:

Since this data repeats every year, this is a periodic curve as shown below:

So, you can note that this is obviously a trigonometric model. Let us try to fit a sine model to this data:

A sine model is of the form: a\sin(b(x-c)) +d

1. We know the data repeats every 12 months. Hence, the period is 12. This gives the value of b to  be b = \frac{2\pi}{12}=\frac{\pi}{6}
2. Since the maximum is 26 °C and the minimum is 5°C, half the distance between these two values will give the amplitude a. Hence a = \dfrac{26-5}{2} = 10.5.
3. The average of the maximum and the minimum points gives the axis of the curve, which is d in the model. So, d = \dfrac{26+5}{2} = 15.5
4. In order to find c in the model, we need to look into where the data crosses the axis (15.5) for the first time. From the table (or from the graph), you can see that it occurs between April and May: Hence, c = \dfrac{4+5}{2} = 4.5

Combining all the above, the model is:

T = 10.5 \sin(\frac{\pi}{6}(x-4.5))+15.5

Graphing this model over the data set, you can see that it fits almost all the points.

 

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