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# The trig functions for any angle and applications

## The trig functions for any angle

So far we have defined the trig functions using right angled triangles. We can now extend these definitions to any angle. We do this by noting that the definitions do not rely on the lengths of the sides of the triangle, but only on the angle. So if we plot any point on the Cartesian plane and then draw a line from the origin to that point, we can work out the angle of that line. In Figure 1 points P and Q have been plotted. A line from the origin to each point is drawn. The dotted lines show how we can construct right angle triangles for each point. Now we can find the angles A and B.

You should find the angle A is 63,4363,43. For angle B, you first work out x (33,6933,69) and then B is 180-33,69=146,31180-33,69=146,31. But what if we wanted to do this without working out these angles and figuring out whether to add or subtract 180 or 90? Can we use the trig functions to do this? Consider point P in Figure 1. To find the angle you would have used one of the trig functions, e.g. tanθtanθ. You should also have noted that the side adjacent to the angle was just the x-co-ordinate and that the side opposite the angle was just the y-co-ordinate. But what about the hypotenuse? Well, you can find that using Pythagoras since you have two sides of a right angled triangle. If we were to draw a circle centered on the origin, then the length from the origin to point P is the radius of the circle, which we denote r. Now we can rewrite all our trig functions in terms of x, y and r. But how does this help us to find angle B? Well, we know that from point Q to the origin is r, and we have the co-ordinates of Q. So we simply use the newly defined trig functions to find angle B! (Try it for yourself and confirm that you get the same answer as before.) One final point to note is that when we go anti-clockwise around the Cartesian plane the angles are positive and when we go clockwise around the Cartesian plane, the angles are negative.

So we get the following definitions for the trig functions:

sin θ = x r cos θ = y r tan θ = y x sin θ = x r cos θ = y r tan θ = y x
(1)

But what if the x- or y-co-ordinate is negative? Do we ignore that, or is there some way to take that into account? The answer is that we do not ignore it. The sign in front of the x- or y-co-ordinate tells us whether or not sin, cos and tan are positive or negative. We divide the Cartesian plane into quadrants and then we can use Figure 2 to tell us whether the trig function is positive or negative. This diagram is known as the CAST diagram.

We can also extend the definitions of the reciprocals in the same way:

cosec θ = r x sec θ = r y cot θ = x y cosec θ = r x sec θ = r y cot θ = x y
(2)

### Exercise 1: Finding the angle

Points R(-1;-3) and point S(3;-3) are plotted in the diagram below. Find the angles αα and ββ.

#### Solution

1. Step 1. Write down what is given and what is required :

We have the co-ordinates of the points R and S. We are required to find two angles. Angle ββ is positive and angle αα is negative.

2. Step 2. Calculate ββ:

We use tan to find ββ, since we are only given x and y. We note that we are in the third quadrant, where tan is positive.

tan ( β ) = y x tan ( β ) = -3-1 β = tan-1 (3) β = 71,57 tan ( β ) = y x tan ( β ) = -3-1 β = tan-1 (3) β = 71,57
(3)
3. Step 3. Calculate αα:

We use tan to calculate αα, since we are only given x and y. We also note that we are in the fourth quadrant, where tan is positive.

tan ( α ) = y x tan ( α ) = -33 α = tan-1 (-1) α = -45 tan ( α ) = y x tan ( α ) = -33 α = tan-1 (-1) α = -45
(4)
4. Step 4. Write the final answer:

Angle αα is -45-45 and angle ββ is 71,5771,57

### Note:

You should observe that in the worked example above, the angle αα is simply the angle that line OS makes with the x-axis. So if you are asked to find out what angle a line makes with the x- or y-axes, then you can use trigonometry!

### Note:

The CAST diagram can be generalized to a very powerful tool for solving trigonometric functions by hand (that is, without a calculator) called the unit circle. You may or may not touch on this in Grade 11 or 12.

## Solving simple trigonometric equations

Using what we have learnt about trig functions we can now solve some simple trig equations. We use the principles from Equations and Inequalities to help us solve trig equations.

### Note:

It is important to note that in general 2sinθsin(2θ)2sinθsin(2θ). In other words doubling (or multiplying by 2) has a different effect from doubling the angle.

### Exercise 2

Solve the following trig equation: 3cos(2x+38)+3=23cos(2x+38)+3=2

#### Solution

1. Step 1. Rearrange the equation:
3cos(2x+38)=2-3 cos(2x+38)=-13 (2x+38)=107,46 2x=107,46-38 2x=69,46 x=34,73 3cos(2x+38)=2-3 cos(2x+38)=-13 (2x+38)=107,46 2x=107,46-38 2x=69,46 x=34,73
(5)
2. Step 2. Write the final answer: x=34,73x=34,73

## Simple Applications of Trigonometric Functions

Trigonometry was probably invented in ancient civilisations to solve practical problems such as building construction and navigating by the stars. In this section we will show how trigonometry can be used to solve some other practical problems.

### Height and Depth

One simple task is to find the height of a building by using trigonometry. We could just use a tape measure lowered from the roof, but this is impractical (and dangerous) for tall buildings. It is much more sensible to measure a distance along the ground and use trigonometry to find the height of the building.

Figure 4 shows a building whose height we do not know. We have walked 100 m away from the building and measured the angle from the ground up to the top of the building. This angle is found to be 38,738,7. We call this angle the angle of elevation. As you can see from Figure 4, we now have a right-angled triangle. As we know the length of one side and an angle, we can calculate the height of the triangle, which is the height of the building we are trying to find.

If we examine the figure, we see that we have the opposite and the adjacent of the angle of elevation and we can write:

tan 38 , 7 = opposite adjacent = height 100 m height = 100 m × tan 38 , 7 = 80 m tan 38 , 7 = opposite adjacent = height 100 m height = 100 m × tan 38 , 7 = 80 m
(6)

#### Exercise 3: Height of a tower

A block of flats is 100m away from a cellphone tower. Someone stands at BB. They measure the angle from BB up to the top of the tower EE to be 62 . This is the angle of elevation. They then measure the angle from BB down to the bottom of the tower at CC to be 34 . This is the angle of depression.What is the height of the cellph one tower correct to 1 decimal place?

##### Solution
1. Step 1. Identify a strategy :

To find the height of the tower, all we have to do is find the length of CDCD and DEDE. We see that BDEBDE and BDCBDC are both right-angled triangles. For each of the triangles, we have an angle and we have the length ADAD. Thus we can calculate the sides of the triangles.

2. Step 2. Calculate CDCD :

We are given that the length ACAC is 100m. CABDCABD is a rectangle so BD=AC=100mBD=AC=100m.

tan ( C B ^ D ) = C D B D C D = B D × tan ( C B ^ D ) = 100 × tan 34 tan ( C B ^ D ) = C D B D C D = B D × tan ( C B ^ D ) = 100 × tan 34
(7)

Use your calculator to find that tan34=0,6745tan34=0,6745. Using this, we find that CD=67,45CD=67,45m

3. Step 3. Calculate DEDE :
tan ( D B ^ E ) = D E B D D E = B D × tan ( D B ^ E ) = 100 × tan 62 = 188 , 07 m tan ( D B ^ E ) = D E B D D E = B D × tan ( D B ^ E ) = 100 × tan 62 = 188 , 07 m
(8)
4. Step 4. Combine the previous answers :

We have that the height of the tower CE=CD+DE=67,45m+188,07m=255.5mCE=CD+DE=67,45m+188,07m=255.5m.

### Maps and Plans

Maps and plans are usually scale drawings. This means that they are an exact copy of the real thing, but are usually smaller. So, only lengths are changed, but all angles are the same. We can use this idea to make use of maps and plans by adding information from the real world.

#### Exercise 4: Scale Drawing

A ship approaching Cape Town Harbour reaches point A on the map, due south of Pretoria and due east of Cape Town. If the distance from Cape Town to Pretoria is 1000km, use trigonometry to find out how far east the ship is from Cape Town, and hence find the scale of the map.

##### Solution
1. Step 1. Identify what happens in the question :

We already know the distance between Cape Town and AA in blocks from the given map (it is 5 blocks). Thus if we work out how many kilometers this same distance is, we can calculate how many kilometers each block represents, and thus we have the scale of the map.

2. Step 2. Identify given information :

Let us denote Cape Town with CC and Pretoria with PP. We can see that triangle APCAPC is a right-angled triangle. Furthermore, we see that the distance ACAC and distance APAP are both 5 blocks. Thus it is an isoceles triangle, and so AC^P=AP^C=45AC^P=AP^C=45.

3. Step 3. Carry out the calculation :
C A = C P × cos ( A C ^ P ) = 1000 × cos ( 45 ) = 1000 2 km C A = C P × cos ( A C ^ P ) = 1000 × cos ( 45 ) = 1000 2 km
(9)

To work out the scale, we see that

5 blocks = 1000 2 km 1 block = 200 2 km 5 blocks = 1000 2 km 1 block = 200 2 km
(10)

#### Exercise 5: Building plan

Mr Nkosi has a garage at his house, and he decides that he wants to add a corrugated iron roof to the side of the garage. The garage is 4m high, and his sheet for the roof is 5m long. If he wants the roof to be at an angle of 55, how high must he build the wall BDBD, which is holding up the roof? Give the answer to 2 decimal places.

##### Solution
1. Step 1. Set out strategy :

We see that the triangle ABCABC is a right-angled triangle. As we have one side and an angle of this triangle, we can calculate ACAC. The height of the wall is then the height of the garage minus ACAC.

2. Step 2. Execute strategy :

If BCBC=5m, and angle AB^C=5AB^C=5, then

A C = B C × sin ( A B ^ C ) = 5 × sin 5 = 5 × 0 , 0871 = 0 . 4358 m A C = B C × sin ( A B ^ C ) = 5 × sin 5 = 5 × 0 , 0871 = 0 . 4358 m
(11)

Thus we have that the height of the wall BD=4m-0.4358m=3.56mBD=4m-0.4358m=3.56m.

#### Applications of Trigonometric Functions

1. A boy flying a kite is standing 30 m from a point directly under the kite. If the string to the kite is 50 m long, find the angle of elevation of the kite.
2. What is the angle of elevation of the sun when a tree 7,15 m tall casts a shadow 10,1 m long?

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