This checkpoint is split into three parts: (1) the basics, (2) the trigonometric ratios and (3) the unit circle.

PART 1 - The Basics

Here are some rules to memorise for the exam when it comes to the basics.

1. The internal angles of a triangle always total 180 degrees.
2. The hypotenuse is the slanted part of a right-angled triangle
3. A complementary angle is an angle, that with others, adds up to 90 degrees (a right angle)
4. A supplementary angle is an angle, that with others, adds up to 180 degrees (a straight angle)

Here’s an example question (similar to your exam)

Find the missing angle in this diagram

a) 6 tan(26)°
b) 34 sin(20°)°
c) 63°
d) 51°

This question is designed to give you far more information than you need in order to confuse your response.

All we really need to know is that the interior angles of a triangle must always add to 180°- the circle, the lengths of the lines etc. are all irrelevant. Since we know that two of the angles are equal to 63°and 54°, we can express the sum of the angles as 63°+54°+ ?=180°. This simplifies to 117°+?=180°, and we can solve for ? by subtracting 117°from both sides to get ?=63°. Our answer is therefore (c).

The most important thing to remember when answering this question is that often more information than you need will be included in order to confuse you. If you know two of the angles in a triangle, you don’t need to use trigonometry or Pythagoras - you can use the fact that the three angles add up to 180°to solve for the last angle.

PART 2 - The Trigonometric Ratios

Trignometry isn’t hard (although some of the language may seem foreign in the beginning) – it’s pretty much governed by a set of rules or ratios to help you calculate parts of a triangle with very little other information.

Here are the rules:

1. a² + b² = c² - this is Pythagoras’ theorem
2. sin(x°)=opposite / hypotenuse(SOH); cos(x°) =adjacent / hypotenuse(CAH); tan(x°)=opposite / adjacent(TOA) – these make up the trigonometric ratios.

Here is a table of values (called values of circular functions). I recommend memorising this table to make it easier for you to solve these types of questions:

Let’s now look at some practice questions…

Question 1
What is the length of the hypotenuse of this triangle in its simplest form?

a) √(116)cm
b) 2√(29)cm
c) √(14)cm
d) 116cm

To answer this question, we need to use Pythagoras’ formula to find the length of the hypotenuse based on the lengths of the other two sides.

We can substitute the lengths of these sides (4cm and 10cm) into the formula a² + b² = c² to get 4² + 10² = c². Since 4 x 4=16 and 10 x 10=100, c² =116. This means that c is equal to the square root of 116 or √(116).

However, this question tries to trick us by offering the answer √(116) (option a) despite asking us to give the solution “in its simplest form.” Therefore, we need to simplify this surd (surds are in checkpoint 12) by removing a square factor (in this case, 2). Since 116 = 4 x 29, we can rewrite this surd as √(4)√(29) which is equivalent to 2√(29). Therefore, our answer is (b).

To recap:
1. Substitute the side lengths of the triangle into Pythagoras’ formula
2. Simplify the result by removing a square factor

Question 2

a° = ?

a) 10°
b) 30°
c) 45°
d) 60°

In this question we know the length of two sides, and we want to know the angle a°. To find this angle, we’ll need to use one of the trigonometric ratios:

• sin(x°)=opposite/hypotenuse(SOH)
• cos(x°) =adjacent/hypotenuse(CAH)
• tan(x°)=opposite/adjacent(TOA)

Where the hypotenuse is the long, diagonal side; the adjacent is the shorter side that borders the angle; and the opposite is the shorter side furthest away from the angle. In this question, we know the length of the two shorter sides (the adjacent and the opposite), meaning that we need to use the ratio tan.

If we substitute the lengths of the two sides into the equation, we know that tan(a°)=5.6 ÷ 5.6, meaning that tan(a°) = 1. In order to find the angle a° we therefore need to find an angle for which tan = 1. We can refer to the table of exact values for circular functions to find this value:

From this table, we can see that the only value of a between 0° and 90°for which tan = 1 is a=45°. Another reason for this is that since tan(x)=sin(x) ÷ cos(x), for tan(x) to equal 1 it must mean that sin(x) = cos(x). For that to be true, the angle a° must be halfway between 0°and 90°- an angle of 45°. Our answer is (c). To recap:

1. Identify the correct trigonometric ratio, following the pattern SOH CAH TOA
2. Substitute in any known angles and side lengths
3. Solve for the unknown value using the exact values of trigonometric ratios

Question 3

If sin(2x)=1, what is the smallest possible value of x?
a) 45°
b) 50°
c) 0°
d) 30°

This question asks us to solve a trigonometric equation for x, using the exact values of sin to find the smallest possible value of x.

First, we need to find the smallest value of a° for which sin(a°)=1, where 2x=a. This is because we cannot just divide both sides by 2 to get sin(x)=1/2, since we cannot divide by a number which is within the trigonometric function.
By referring to the table of exact values of sin, we know that the smallest possible value of a is 90°.
If 2x=90°, we can divide both sides by 2 to find that x=45°. Therefore, our answer is (a). To recap:

1. Find the smallest value for which sin(a°)=1
2. Substitute in a=2x

Question 4
What is the perimeter of this triangle?

a) 9+3√3m
b) 12m
c) 3√3m
d) 9m
This question asks us to find the perimeter of the triangle, meaning that we need to find the length of each of its sides. We know the length of one side and its adjacent angle, meaning that we can use trigonometric ratios to find the length of another side.

We can use either cos (to find the hypotenuse) or tan (to find the other shorter side). In this case, we’ll use tan to find that tan(60°)=opposite/3. We know that tan(60°)=√3, meaning that √3=opposite/3. If we multiply both sides by 3, we find that the length of the side on the bottom is 3√3m.

Next, we need to find the length of the hypotenuse. We can do this either by using Pythagoras’ formula (a² + b² = c²) or by using another trigonometric ratio (either sin or cos). In this case, it’s probably easier to substitute the lengths of the shorter sides into Pythagoras’ formula to get (3)² + (3√3)² = c², where c is the length of the hypotenuse. By simplifying this equation using 3² = 9 and √(3)² = 3, we get a result of 36=c². Therefore c=√36=6.

We now know the lengths of all sides (expressed in diagram below) and can add them together to find the perimeter of the shape. 3+6+3√3=9+3√3, meaning that our answer is (a).

To recap:

1. Find the lengths of all sides using either Pythagoras or trigonometric ratios (SOH CAH TOA)
2. Add together the side lengths to find the perimeter

PART 3 - The Unit Circle

Here’s what you need to know about the Unit Circle:

1. A unit circle is a circle with a radius (line half circle) of 1 and because it has a radius of 1, you can measure sin, cos and tan with it.
2. The "sides" can be positive or negative according to the rules of Cartesian coordinates. This makes the sine, cosine and tangent change between positive and negative values also.
3. There are additional ratios for you to remember: cosec or csc A = hypotenuse / opposite, sec A = hypotenuse / adjacent and cotan/cot/cotg/ctg/ctn A = adjacent / opposite
4. There is another unit of measurement and that is called radians (rad / r) and you can use that measurement to show the circumference of a circle (2𝜋r).
   
5. Degree to radians = divide by 180 then times by 𝜋. Radians to degrees is the reverse, divide by 𝜋 and then multiply by 180.
6. The table showing when sin, cos and tan are positive or negative based on the unit circle.

By Jim.belk - Own work, CC BY-SA 3.0, Link

That’s a lot to take in! Let’s see it now in action…

Question 1

sin(30°)-cos(60°)= ?
a) tan(0°)
b) sin(0°)
c) -30°
d) 0
This question asks us to find the value of sin(30°) and cos(60°), and then subtract one from the other.

We know that 30° and 60°are both 30°away from the borders of their quadrant (in other words, 0 + 30° = 30° and 90°- 60°= 30°), and so their coordinates on the unit circle are the inverse of each other. That means that if the coordinates of 30° are (a, b), then the coordinates of 60°are (b, a).

We also know that the x-coordinate of a point on the unit circle is cos(a), and the y-coordinate of a point on the unit circle is sin(a). Therefore, if the coordinates of 30° are (cos(30°), sin(30°)), the coordinates of 60° are (sin(30°), cos(30°) OR (cos(60°), sin(60°). This means that cos(60°) is equivalent to sin(30°).

Therefore, we can simplify our equation to sin(30°)-sin(30°), which of course must equal zero (since any number minus itself equals zero). Our answer is (d). To recap:

1. Find the position of the two angles on the unit circle
2. Find their coordinates
3. Solve the equation for the missing value

Question 2

What is the supplementary angle to the angle 6𝜋/8?
a) 42°
b) 2𝜋/8
c) -6𝜋/8
d) 4°

This question asks us to understand angles in radians, such as 6𝜋/8. If it makes you more comfortable, you can convert this angle to degrees by multiplying it by 180/𝜋 to get a value of 135°- however, you can also solve the question entirely in radians.

We know that supplementary angles sum to 180° or 𝜋 radians, so we need to find an angle which, when added to 6𝜋/8, equals 𝜋 radians. We can express this in an equation as 𝜋 =6𝜋/8 + ?. In order to put all values on the same base (8), we can rewrite this as 8𝜋/8=6𝜋/8+?𝜋/8- meaning that ?+6=8.

This means that our unknown value is 2, and therefore the angle is 2𝜋/8. Our answer is (b). To recap:

1. If required, convert all angles in radians to degrees by multiplying by 180/𝜋
2. Express the sum of the angles as an equation equalling 180°or 𝜋
3. Solve for the unknown angle

Question 3

If sin(a°) = -1/√3 and 0 ≤ a ≤ 3𝜋/2, what quadrant is a° in?
a) Quadrant 1
b) Quadrant 2
c) Quadrant 3
d) Quadrant 4

We know that sin(a°)= -1/√3,and we want to know which quadrant of the unit circle the angle a° is in. We know that the domain of a is between 0 and 3𝜋/2, meaning that it must be in the first quadrant (0 - 𝜋/-2), second quadrant (𝜋/2 - 𝜋), or third quadrant (𝜋, 3𝜋/2). Therefore, we can knock out option (d).

Next, we need to think about which quadrants sin is positive in. The following table summarises which values are positive and negative in each quadrant.

From this table, we can see that if sin(a°) is negative, “a” must be in the 3rd or 4th quadrant - but since we have already decided that it can’t be in the 4th quadrant, the answer must be (c). To recap:

1. Look carefully at the excluded values of a°
2. Work out which quadrants sin is negative in

Question 4
Which of these points is on the unit circle?
a) (1, 1)
b) (√(3)/2, 1/2)
c) (14/3, 2)
d) (1/4, 3/4)

To answer this question, we need to use the formula of the unit circle to work out which point is on this line.
The formula of the unit circle is the standard formula of a circle with radius 1 and center (0, 0) - y² + x² = 1. We need to substitute the points in each of these options into that equation to work out if the statement is correct.

(1)² + (1)² = 1, 1 + 1 = 1, ∴ Incorrect
(√(3)/2)² + (1/2)² = 1, 3/4+ 1/4= 1, ∴ Correct
(14/3)² + (2)² = 1, 196/9+ 4 = 1, ∴ Incorrect
(1/4)² + (3/4)² = 1, 1/16+9/16= 1, ∴ Incorrect

The only one of these points on the unit circle is (√(3)/2, 1/2), since it’s the only one which satisfies the equation y² + x² = 1 and this means the answer is (b). To recap:

1. Recall the equation of the unit circle
2. Work out which coordinates fit the equation

Now it's time to do your assignment.

1. Download the assignment questions here.
2. Print it out or if you want to do it electronically, save it.
3. Complete the questions to it.
4. Then check the solutions on the video below. The answer key is also on the final page of your downloaded assignment questions.

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