Answer key · Unit 8

Trigonometry.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Sine, cosine and tangent

In a right triangle, the ratio of two sides depends only on the angle. These three ratios are called sine, cosine and tangent.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Naming the sides
Practice · Choose
Question
The board names the sides as seen from angle $A$ ⁠. Now look from angle $C$ ⁠, at the top corner. Which side is the side adjacent to $C$ ⁠?

Correct answer
- Correct $BC$ ⁠
Setup, choices and feedback
The Scene at this step
Angle A
35 °fixed
Side AB
6fixed
Given / Worked out
Worked outfixed
Other settings
Trig/nodes/board/props/trig = "names"
Choices
Shown in a random order.
- Correct $BC$ ⁠
- Wrong $AB$ ⁠
  Misconception: Thinks 'opposite' and 'adjacent' are fixed names of two sides, instead of depending on the angle you look from.
  Feedback from the catalogue
  “Opposite” and “adjacent” depend on the angle you look from. The leg that touches the angle is the side adjacent to it, and the other leg is opposite it. Move to the other angle and the two legs swap roles. Only the hypotenuse stays the hypotenuse: it is always opposite the right angle.
- Wrong $CA$ ⁠
  Feedback
  $CA$ ⁠ does touch $C$ ⁠, but it is opposite the right angle: it is the hypotenuse, whichever angle you look from.
Feedback for a right answer
$BC$ ⁠ leaves $C$ ⁠ and is not the hypotenuse, so it is the side adjacent to $C$ ⁠. Seen from $A$ ⁠, the same side is the opposite one. The names belong to the angle, not to the side.

Hint
The side adjacent to an angle is a leg that touches it. Which two sides leave $C$ ⁠, and which of them is not the hypotenuse?

Worked solution
1. Two sides leave $C$ ⁠: $CA$ ⁠ and $CB$ ⁠.
2. $CA$ ⁠ is opposite the right angle at $B$ ⁠, so it is the hypotenuse.
3. The other side leaving $C$ ⁠, $BC$ ⁠, is the side adjacent to $C$ ⁠.
4. The remaining leg, $AB$ ⁠, does not touch $C$ ⁠: it is the side opposite $C$ ⁠.
If you chose $AB$ ⁠, you kept the name it had from $A$ ⁠. When you move to the other angle, the two legs swap names.
Step 2 of 5
Open this step
Same angle, same ratio
Practice · Predict, then look
Question
$A = 30^\circ$ ⁠ and $AB = 4$ ⁠. When $AB$ ⁠ grows to $8$ ⁠, the whole triangle doubles in size and angle $A$ ⁠ does not change. What happens to $\sin A$ ⁠, the opposite side divided by the hypotenuse?

Correct answer
- Correct It stays $0.5$ ⁠
Setup, choices and feedback
The Scene at this step
Angle A
30 °fixed
Side AB
4
Given / Worked out
Worked outfixed
Other settings
Trig/nodes/board/props/trig = "sin"
Choices
- Correct It stays $0.5$ ⁠
- Wrong It doubles, to $1$ ⁠
  Misconception: Treats sin, cos or tan as a length (a side), not as a ratio of two sides.
  Feedback from the catalogue
  A sine is not a side but a ratio: one side divided by another. When the triangle doubles in size, the opposite side and the hypotenuse both double, and their ratio does not change. So $\sin A$ ⁠ depends only on the angle, not on the size of the triangle.
- Wrong It halves, to $0.25$ ⁠
  Feedback
  Both the opposite side and the hypotenuse grow, so there is no reason for the ratio to shrink: each of them is multiplied by the same number.
Once the prediction is locked in, the Scene changes
Side AB
8
The Scene then shows: 0.5 · Triangle: sin A; with a right angle at B, the side opposite A (BC) over the hypotenuse (CA)/nodes/board/outputs/sinA
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The opposite side grows from $2.31$ ⁠ to $4.62$ ⁠ and the hypotenuse from $4.62$ ⁠ to $9.24$ ⁠, and the ratio stays $0.5$ ⁠. In a triangle with a $30^\circ$ ⁠ angle the side opposite it is always half the hypotenuse: $\sin 30^\circ = 0.5$ ⁠.

Worked solution
1. When every side is multiplied by $2$ ⁠, the new triangle is similar to the old one: the same angles.
2. $\sin A = \frac{BC}{CA}$ ⁠, and the top and the bottom are both multiplied by $2$ ⁠.
3. $\frac{2 \cdot BC}{2 \cdot CA} = \frac{BC}{CA}$ ⁠: the ratio does not change.
4. So $\sin 30^\circ = 0.5$ ⁠ in every right triangle with a $30^\circ$ ⁠ angle.
If you chose that it doubles, you treated the sine as the length of a side. It is a ratio of two sides.
Step 3 of 5
Open this step
Working out a sine
Practice · Calculate
Question
$A = 40^\circ$ ⁠. The side opposite $A$ ⁠ is $5.03$ ⁠ and the hypotenuse is $7.83$ ⁠. What is $\sin 40^\circ$ ⁠? Give two decimal places.

Correct answer
0.6428 ± 0.006
Setup, choices and feedback
The Scene at this step
Angle A
40 °fixed
Side AB
6fixed
Given / Worked out
Given
Other settings
Trig/nodes/board/props/trig = "sin"
Other settings
Unknown/nodes/board/props/unknown = "sinA"
Accepted within
± 0.006
Computed from the Scene at this step’s setting
Triangle: sin A; with a right angle at B, the side opposite A (BC) over the hypotenuse (CA)/nodes/board/outputs/sinA
Wrong answers it recognises
- 5.035 ± 0.05
  Misconception: Treats sin, cos or tan as a length (a side), not as a ratio of two sides.
  Feedback from the catalogue
  A sine is not a side but a ratio: one side divided by another. When the triangle doubles in size, the opposite side and the hypotenuse both double, and their ratio does not change. So $\sin A$ ⁠ depends only on the angle, not on the size of the triangle.
- 0.766 ± 0.006
  Misconception: Mixes up the opposite and the adjacent side, and so uses cos for sin or the reverse.
  Feedback from the catalogue
  The side opposite an angle does not touch it; the side adjacent to an angle is one of the two sides that make it. Sine uses the opposite side, cosine the adjacent one. Label all three sides before you pick a ratio.
- 0.8391 ± 0.006
  Misconception: Picks sin, cos or tan without checking which two sides the question links.
  Feedback from the catalogue
  Before you pick a ratio, ask which two sides are involved: the one you know and the one you want. Opposite and hypotenuse: sine. Adjacent and hypotenuse: cosine. Opposite and adjacent: tangent.
Feedback for a right answer
$\frac{5.03}{7.83} \approx 0.64$ ⁠. That is $\sin 40^\circ$ ⁠ in every right triangle, large or small. A calculator in degrees gives $0.6428$ ⁠.

Hint
Sine: the side opposite the angle divided by the hypotenuse.

Worked solution
1. The sine is the side opposite the angle divided by the hypotenuse: $\sin A = \frac{BC}{CA}$ ⁠.
2. From the board: $BC = 5.03$ ⁠ and $CA = 7.83$ ⁠.
3. $\sin 40^\circ = \frac{5.03}{7.83} \approx 0.64$ ⁠.
If you got $0.77$ ⁠, you divided the side adjacent to $A$ ⁠, $6$ ⁠, by the hypotenuse: that is the cosine. If you answered $5.03$ ⁠, you gave the side itself, not the ratio.
Step 4 of 5
Open this step
Cosine
Practice · Reach the target
Question
Side AB is locked. Set Angle A so that $\cos A = 0.5$ ⁠, that is, the side adjacent to $A$ ⁠ is exactly half the hypotenuse, and press Check.

Goal
- 0.5 ± 0.005
  At the start of the step: 0.866
Setup, choices and feedback
The Scene at this step
Angle A
30 °
Side AB
6fixed
Given / Worked out
Worked outfixed
Other settings
Trig/nodes/board/props/trig = "cos"
Read from the Scene at
Triangle: cos A; with a right angle at B, the side adjacent to A (AB) over the hypotenuse (CA)/nodes/board/outputs/cosA
One setting that reaches it
Angle A
60 °
Feedback for a right answer
$\cos 60^\circ = 0.5$ ⁠. It is the triangle from the last step, turned round: the side opposite $30^\circ$ ⁠ is the side adjacent to $60^\circ$ ⁠, so $\cos 60^\circ = \sin 30^\circ$ ⁠.

Feedback for a wrong answer
Not yet. As the angle grows the cosine shrinks: the bottom line shows it at every moment.

Hint
At $30^\circ$ ⁠ the cosine is $0.866$ ⁠. To make it smaller, the angle has to grow.

Worked solution
1. $\cos A = \frac{AB}{CA}$ ⁠, and it has to be $0.5$ ⁠: the hypotenuse twice as long as the side adjacent to $A$ ⁠.
2. At $A = 30^\circ$ ⁠ the cosine is $0.866$ ⁠, too big. As the angle grows, the triangle stands up and the cosine shrinks.
3. At $60^\circ$ ⁠ the other angle is $30^\circ$ ⁠, and the side adjacent to $A$ ⁠ is the side opposite the $30^\circ$ ⁠ angle: half the hypotenuse.
4. Set Angle A to $60^\circ$ ⁠.
Step 5 of 5
Open this step
Tangent
Practice · Predict, then look
Question
$\tan A$ ⁠ is the side opposite $A$ ⁠ divided by the side adjacent to $A$ ⁠. What will $\tan 45^\circ$ ⁠ be?

Correct answer
- Correct $1$ ⁠
Setup, choices and feedback
The Scene at this step
Angle A
30 °fixed
Side AB
6fixed
Given / Worked out
Worked outfixed
Other settings
Trig/nodes/board/props/trig = "tan"
Choices
- Correct $1$ ⁠
- Wrong $0.5$ ⁠
  Feedback
  $45^\circ$ ⁠ is half a right angle, but the ratio of the sides is not a half. Think about what you know of the other two angles in the triangle.
- Wrong $0.707$ ⁠
  Misconception: Picks sin, cos or tan without checking which two sides the question links.
  Feedback from the catalogue
  Before you pick a ratio, ask which two sides are involved: the one you know and the one you want. Opposite and hypotenuse: sine. Adjacent and hypotenuse: cosine. Opposite and adjacent: tangent.
- Wrong $45$ ⁠
  Misconception: Confuses an angle with its sine, cosine or tangent.
  Feedback from the catalogue
  The angle is measured in degrees, and its sine, cosine and tangent are ratios of sides. $\sin 30^\circ = 0.5$ ⁠: from $30^\circ$ ⁠ you get to $0.5$ ⁠, and from $0.5$ ⁠ you get back to $30^\circ$ ⁠ with the inverse function, $\sin^{-1}$ ⁠.
Once the prediction is locked in, the Scene changes
Angle A
45 °fixed
The Scene then shows: 1 · Triangle: tan A; with a right angle at B, BC over AB/nodes/board/outputs/tanA
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$A = 45^\circ$ ⁠ and $B = 90^\circ$ ⁠, so $C = 45^\circ$ ⁠ too. A triangle with two equal angles is isosceles: the two legs are equal, $6$ ⁠ and $6$ ⁠, and their ratio is $1$ ⁠.

Worked solution
1. The angles add up to $180^\circ$ ⁠: $C = 180^\circ - 90^\circ - 45^\circ = 45^\circ$ ⁠.
2. In a triangle with two equal angles, the sides opposite them are equal: $BC = AB$ ⁠.
3. $\tan 45^\circ = \frac{BC}{AB} = 1$ ⁠.
If you chose $0.707$ ⁠, you worked out $\sin 45^\circ$ ⁠ or $\cos 45^\circ$ ⁠, with the hypotenuse. The tangent uses the two legs.

Lesson 2 of 4 · About 25 minutes

Finding sides and angles

One angle and one side are enough to work out every other side of a right triangle, and two sides are enough to work out the angle: a ladder, a ramp, a tree, and inverse functions.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A ladder against a wall
Practice · Calculate
Question
A ladder 10 m long leans against a wall at $60^\circ$ ⁠ to the ground. How high up the wall does it reach, in metres?

Correct answer
8.66 ± 0.05
Setup, choices and feedback
The Scene at this step
Angle A
60 °fixed
Side AB
5fixed
Given / Worked out
Given
Other settings
Trig/nodes/board/props/trig = "sin"
Other settings
Unknown/nodes/board/props/unknown = "sideA"
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Triangle: BC, opposite A/nodes/board/outputs/sideA
Wrong answers it recognises
- 5 ± 0.05
  Misconception: Mixes up the opposite and the adjacent side, and so uses cos for sin or the reverse.
  Feedback from the catalogue
  The side opposite an angle does not touch it; the side adjacent to an angle is one of the two sides that make it. Sine uses the opposite side, cosine the adjacent one. Label all three sides before you pick a ratio.
- 0.866 ± 0.01
  Misconception: Gives the ratio itself as the answer, forgetting to multiply by the known side.
  Feedback from the catalogue
  $\sin A$ ⁠ is only the ratio, a number without units. To get a length, multiply it by the known side: if $\sin A = \frac{x}{c}$ ⁠, then $x = c \cdot \sin A$ ⁠.
- 11.55 ± 0.05
  Misconception: Divides the known side by the ratio when it should multiply (or the reverse).
  Feedback from the catalogue
  Write the ratio with the unknown in its place, and only then rearrange. When the known side is on the bottom, multiply by it: $\frac{x}{10} = 0.866$ ⁠ gives $x = 10 \cdot 0.866$ ⁠. When the unknown is on the bottom, divide.
Feedback for a right answer
$10 \cdot \sin 60^\circ = 10 \cdot 0.866 \approx 8.66$ ⁠ metres. Pythagoras checks it: $\sqrt{10^2 - 5^2} = \sqrt{75} \approx 8.66$ ⁠.

Hint
The height is the side opposite the angle, and the ladder is the hypotenuse. Which ratio links them?

Worked solution
1. The ladder is the hypotenuse, $10$ ⁠, and the height on the wall is the side opposite the $60^\circ$ ⁠ angle.
2. Opposite and hypotenuse: sine. $\sin 60^\circ = \frac{x}{10}$ ⁠.
3. Multiply by $10$ ⁠: $x = 10 \cdot \sin 60^\circ$ ⁠.
4. $\sin 60^\circ \approx 0.866$ ⁠, so $x \approx 8.66$ ⁠ metres.
If you got $5$ ⁠, you used the cosine: that is how far the foot of the ladder is from the wall. If you got $0.866$ ⁠, you stopped at the ratio and forgot to multiply by $10$ ⁠.
Step 2 of 5
Open this step
A ramp
Practice · Calculate
Question
A ramp 12 m long rises at $20^\circ$ ⁠. How long is it along the ground, side $AB$ ⁠, in metres? ( $\cos 20^\circ \approx 0.940$ ⁠)

Correct answer
11.28 ± 0.05
Setup, choices and feedback
The Scene at this step
Angle A
20 °fixed
Side AB
11.28hidden
Given / Worked out
Given
Other settings
Trig/nodes/board/props/trig = "cos"
Other settings
Unknown/nodes/board/props/unknown = "sideC"
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Triangle: AB = base, opposite C/nodes/board/outputs/sideC
Wrong answers it recognises
- 4.104 ± 0.05
  Misconception: Mixes up the opposite and the adjacent side, and so uses cos for sin or the reverse.
  Feedback from the catalogue
  The side opposite an angle does not touch it; the side adjacent to an angle is one of the two sides that make it. Sine uses the opposite side, cosine the adjacent one. Label all three sides before you pick a ratio.
- 12.77 ± 0.05
  Misconception: Divides the known side by the ratio when it should multiply (or the reverse).
  Feedback from the catalogue
  Write the ratio with the unknown in its place, and only then rearrange. When the known side is on the bottom, multiply by it: $\frac{x}{10} = 0.866$ ⁠ gives $x = 10 \cdot 0.866$ ⁠. When the unknown is on the bottom, divide.
Feedback for a right answer
$12 \cdot \cos 20^\circ \approx 12 \cdot 0.940 \approx 11.3$ ⁠ metres. At a small angle the adjacent side is nearly as long as the hypotenuse.

Hint
The ground is the side adjacent to the angle, and the ramp is the hypotenuse.

Worked solution
1. The ramp is the hypotenuse, $12$ ⁠. The length on the ground is the side adjacent to the $20^\circ$ ⁠ angle.
2. Adjacent and hypotenuse: cosine. $\cos 20^\circ = \frac{x}{12}$ ⁠.
3. $x = 12 \cdot \cos 20^\circ \approx 12 \cdot 0.940 \approx 11.3$ ⁠ metres.
If you got $4.1$ ⁠, you used the sine: that is the height the ramp climbs, the side opposite the angle.
Step 3 of 5
Open this step
The height of a tree
Practice · Calculate
Question
You stand 15 m from a tree and see its top at $35^\circ$ ⁠ above the ground. How tall is the tree, in metres? ( $\tan 35^\circ \approx 0.700$ ⁠)

Correct answer
10.5 ± 0.06
Setup, choices and feedback
The Scene at this step
Angle A
35 °fixed
Side AB
15fixed
Given / Worked out
Given
Other settings
Trig/nodes/board/props/trig = "tan"
Other settings
Unknown/nodes/board/props/unknown = "sideA"
Accepted within
± 0.06
Computed from the Scene at this step’s setting
Triangle: BC, opposite A/nodes/board/outputs/sideA
Wrong answers it recognises
- 8.6 ± 0.05
  Misconception: Picks sin, cos or tan without checking which two sides the question links.
  Feedback from the catalogue
  Before you pick a ratio, ask which two sides are involved: the one you know and the one you want. Opposite and hypotenuse: sine. Adjacent and hypotenuse: cosine. Opposite and adjacent: tangent.
- 21.4 ± 0.1
  Misconception: Divides the known side by the ratio when it should multiply (or the reverse).
  Feedback from the catalogue
  Write the ratio with the unknown in its place, and only then rearrange. When the known side is on the bottom, multiply by it: $\frac{x}{10} = 0.866$ ⁠ gives $x = 10 \cdot 0.866$ ⁠. When the unknown is on the bottom, divide.
Feedback for a right answer
$15 \cdot \tan 35^\circ \approx 15 \cdot 0.700 \approx 10.5$ ⁠ metres. No climbing, and no hypotenuse needed.

Hint
The distance is the side adjacent to the angle, and the height is the side opposite it. Which ratio uses the two legs?

Worked solution
1. The distance from the tree, $15$ ⁠, is the side adjacent to the angle. The height is the side opposite it.
2. Opposite and adjacent: tangent. $\tan 35^\circ = \frac{h}{15}$ ⁠.
3. $h = 15 \cdot \tan 35^\circ \approx 15 \cdot 0.700 \approx 10.5$ ⁠ metres.
If you got $8.6$ ⁠, you used the sine, as if $15$ ⁠ were the hypotenuse. If you got $21.4$ ⁠, you divided by $0.700$ ⁠ instead of multiplying.
Step 4 of 5
Open this step
From a ratio to an angle
Practice · Reach the target
Question
Side AB is locked at $4$ ⁠. Set Angle A so that the side opposite $A$ ⁠ is $3$ ⁠, and press Check.

Goal
- 3 ± 0.05
  At the start of the step: 2.309
Setup, choices and feedback
The Scene at this step
Angle A
30 °
Side AB
4fixed
Given / Worked out
Worked outfixed
Other settings
Trig/nodes/board/props/trig = "tan"
Read from the Scene at
Triangle: BC, opposite A/nodes/board/outputs/sideA
One setting that reaches it
Angle A
37 °
Feedback for a right answer
$37^\circ$ ⁠ gives $3.01$ ⁠, as close as Angle A can get. The exact angle is $\tan^{-1} 0.75 \approx 36.87^\circ$ ⁠. It is the $3$ ⁠, $4$ ⁠, $5$ ⁠ triangle from the geometry Unit.

Feedback for a wrong answer
Not yet. As the angle grows, the side opposite it gets longer: the bottom line shows the tangent.

Hint
You need $\tan A = \frac{3}{4} = 0.75$ ⁠. At $30^\circ$ ⁠ the tangent is $0.577$ ⁠, at $45^\circ$ ⁠ it is $1$ ⁠.

Worked solution
1. The side adjacent to $A$ ⁠ is $4$ ⁠, and the side opposite $A$ ⁠ has to be $3$ ⁠.
2. Opposite and adjacent: $\tan A = \frac{3}{4} = 0.75$ ⁠.
3. $0.75$ ⁠ lies between $\tan 30^\circ \approx 0.577$ ⁠ and $\tan 45^\circ = 1$ ⁠, so the angle lies between $30^\circ$ ⁠ and $45^\circ$ ⁠.
4. At $37^\circ$ ⁠: $4 \cdot \tan 37^\circ \approx 3.01$ ⁠. The exact angle is $\tan^{-1} 0.75 \approx 36.87^\circ$ ⁠.
Step 5 of 5
Open this step
The inverse function
Practice · Calculate
Question
In a right triangle, the side opposite $A$ ⁠ is $5$ ⁠ and the hypotenuse is $10$ ⁠. How many degrees is angle $A$ ⁠?

Correct answer
30 ± 0.5
Setup, choices and feedback
The Scene at this step
Angle A
30 °hidden
Side AB
8.66hidden
Given / Worked out
Given
Other settings
Trig/nodes/board/props/trig = "sin"
Other settings
Show/nodes/board/props/show = "sides"
Other settings
Unknown/nodes/board/props/unknown = "angleA"
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Triangle: the angle at A/nodes/board/outputs/angleA
Wrong answers it recognises
- 0.5 ± 0.01
  Misconception: Confuses an angle with its sine, cosine or tangent.
  Feedback from the catalogue
  The angle is measured in degrees, and its sine, cosine and tangent are ratios of sides. $\sin 30^\circ = 0.5$ ⁠: from $30^\circ$ ⁠ you get to $0.5$ ⁠, and from $0.5$ ⁠ you get back to $30^\circ$ ⁠ with the inverse function, $\sin^{-1}$ ⁠.
- 60 ± 0.5
  Misconception: Mixes up the opposite and the adjacent side, and so uses cos for sin or the reverse.
  Feedback from the catalogue
  The side opposite an angle does not touch it; the side adjacent to an angle is one of the two sides that make it. Sine uses the opposite side, cosine the adjacent one. Label all three sides before you pick a ratio.
Feedback for a right answer
$\sin A = \frac{5}{10} = 0.5$ ⁠, and $\sin^{-1} 0.5 = 30^\circ$ ⁠. Choose Worked out under Look inside to see the angle.

Hint
Work out $\sin A$ ⁠ first. Which angle has that sine? You met it in the last Lesson.

Worked solution
1. Opposite and hypotenuse: $\sin A = \frac{5}{10} = 0.5$ ⁠.
2. You want the angle whose sine is $0.5$ ⁠: $A = \sin^{-1} 0.5$ ⁠.
3. In the last Lesson you saw that $\sin 30^\circ = 0.5$ ⁠, so $A = 30^\circ$ ⁠.
If you answered $0.5$ ⁠, you stopped at the ratio: that is the sine, not the angle. If you answered $60^\circ$ ⁠, you used $\cos^{-1}$ ⁠, as if $5$ ⁠ were the side adjacent to $A$ ⁠: $60^\circ$ ⁠ is angle $C$ ⁠.

Lesson 3 of 4 · About 30 minutes

The unit circle and radians

Sine and cosine for every angle, above 90 degrees too: the height and width of a point on a circle of radius 1. And radians, the other way to measure angles.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
More than 90 degrees
Practice · Predict, then look
Question
At $30^\circ$ ⁠ the bead is at height $0.5$ ⁠, so $\sin 30^\circ = 0.5$ ⁠. What will $\sin 150^\circ$ ⁠ be?

Correct answer
- Correct $0.5$ ⁠
Setup, choices and feedback
The Scene at this step
Angle
30 °fixed
Circle / Waves
Circle
Choices
- Correct $0.5$ ⁠
- Wrong $-0.5$ ⁠
  Misconception: Thinks sine and cosine are never negative, as if they were lengths.
  Feedback
  At $150^\circ$ ⁠ the bead is still above the horizontal axis, in the second quadrant: its height is positive. Only its width has turned negative.
- Wrong It has none: no right triangle has a $150^\circ$ ⁠ angle
  Misconception: Thinks sine and cosine exist only for angles below 90°, because a right triangle has no larger angle.
  Feedback from the catalogue
  On the unit circle the sine is the height of the point and the cosine its width, so they have a value for every angle, above $90^\circ$ ⁠ too. For acute angles that is exactly the ratio in the triangle; beyond them the definition carries on.
- Wrong $1.5$ ⁠
  Feedback
  The height of a point on a circle of radius $1$ ⁠ cannot be more than $1$ ⁠.
Once the prediction is locked in, the Scene changes
Angle
150 °fixed
The Scene then shows: 0.5 · Unit-circle: sin θ, the bead's height/nodes/board/outputs/sin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$150^\circ$ ⁠ is $30^\circ$ ⁠ short of half a turn: the mirror image of $30^\circ$ ⁠ across the vertical axis. The height is the same, $0.5$ ⁠, and the width has flipped: $\cos 150^\circ \approx -0.866$ ⁠.

Worked solution
1. On the unit circle, $\sin\theta$ ⁠ is the height of the bead.
2. $150^\circ = 180^\circ - 30^\circ$ ⁠: the bead is in the second quadrant, $30^\circ$ ⁠ above the left half of the horizontal axis.
3. That is the mirror image of the point at $30^\circ$ ⁠ across the vertical axis: the same height, the opposite width.
4. So $\sin 150^\circ = \sin 30^\circ = 0.5$ ⁠.
If you chose that it has none, you were thinking only of a triangle. On the unit circle every angle has a sine: the height of the bead.
Step 2 of 7
Open this step
Signs in the quadrants
Practice · Predict, then look
Question
The bead will turn from $60^\circ$ ⁠ to $240^\circ$ ⁠. What signs will the sine and the cosine have there?

Correct answer
- Correct $\sin$ ⁠ negative and $\cos$ ⁠ negative
Setup, choices and feedback
The Scene at this step
Angle
60 °fixed
Circle / Waves
Circle
Choices
- Correct $\sin$ ⁠ negative and $\cos$ ⁠ negative
- Wrong $\sin$ ⁠ positive and $\cos$ ⁠ positive
  Misconception: Thinks sine and cosine are never negative, as if they were lengths.
  Feedback from the catalogue
  The sine is the height of the point on the unit circle and the cosine its width, like coordinates. Below the horizontal axis the height is negative, and left of the vertical axis the width is negative.
- Wrong $\sin$ ⁠ negative and $\cos$ ⁠ positive
  Feedback
  That is the fourth quadrant, between $270^\circ$ ⁠ and $360^\circ$ ⁠: down and to the right. $240^\circ$ ⁠ has not got there yet.
- Wrong $\sin$ ⁠ positive and $\cos$ ⁠ negative
  Feedback
  That is the second quadrant, between $90^\circ$ ⁠ and $180^\circ$ ⁠: up and to the left. $240^\circ$ ⁠ is already past half a turn.
Once the prediction is locked in, the Scene changes
Angle
240 °fixed
The Scene then shows: -0.866 · Unit-circle: sin θ, the bead's height/nodes/board/outputs/sin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$240^\circ$ ⁠ is $60^\circ$ ⁠ past half a turn, in the third quadrant: down and to the left. $\sin 240^\circ \approx -0.866$ ⁠ and $\cos 240^\circ = -0.5$ ⁠, the same numbers as at $60^\circ$ ⁠, with a minus.

Worked solution
1. $240^\circ = 180^\circ + 60^\circ$ ⁠: half a turn and $60^\circ$ ⁠ more.
2. After half a turn the bead is on the left, below the horizontal axis: the third quadrant.
3. Below the horizontal axis the height is negative: $\sin 240^\circ < 0$ ⁠.
4. Left of the vertical axis the width is negative: $\cos 240^\circ < 0$ ⁠.
If you chose both positive, you treated the sine and the cosine as lengths. On the unit circle they are coordinates, and can be negative.
Step 3 of 7
Open this step
Two angles, one sine
Practice · Reach the target
Question
Find an angle in the first turn, other than $30^\circ$ ⁠, where $\sin\theta = 0.5$ ⁠ too. Set Angle and press Check.

Goal
- 0.5 ± 0.005
  At the start of the step: 1
- -0.866 ± 0.01
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Angle
90 °
Circle / Waves
Circle
Read from the Scene at
Unit-circle: sin θ, the bead's height/nodes/board/outputs/sin
Read from the Scene at
Unit-circle: cos θ, the bead's width/nodes/board/outputs/cos
One setting that reaches it
Angle
150 °
Feedback for a right answer
$150^\circ = 180^\circ - 30^\circ$ ⁠. Every height between $-1$ ⁠ and $1$ ⁠ has two angles in one turn, which is why $\sin^{-1} 0.5$ ⁠ on a calculator gives only one of them, $30^\circ$ ⁠.

Feedback for a wrong answer
Not yet. You need the point at height $0.5$ ⁠ on the left of the circle, where the cosine is negative.

Hint
A horizontal line at height $0.5$ ⁠ crosses the circle once on the right and once on the left. The left point is the mirror image of $30^\circ$ ⁠.

Worked solution
1. $\sin\theta = 0.5$ ⁠ means the bead is at height $0.5$ ⁠.
2. A horizontal line at height $0.5$ ⁠ crosses the circle at two points: at $30^\circ$ ⁠, and at its mirror image across the vertical axis.
3. The mirror image is $30^\circ$ ⁠ short of half a turn: $180^\circ - 30^\circ = 150^\circ$ ⁠.
4. Set Angle to $150^\circ$ ⁠.
Step 4 of 7
Open this step
Double the angle
Practice · Predict, then look
Question
$\sin 30^\circ = 0.5$ ⁠. What will $\sin 60^\circ$ ⁠ be?

Correct answer
- Correct About $0.866$ ⁠: more than $0.5$ ⁠, less than $1$ ⁠
Setup, choices and feedback
The Scene at this step
Angle
30 °fixed
Circle / Waves
Circle
Choices
- Wrong $1$ ⁠: twice as much
  Misconception: Thinks doubling the angle doubles the sine: sin 2x = 2 sin x.
  Feedback from the catalogue
  The sine does not grow at a steady rate with the angle: the height on the circle rises fast near $0^\circ$ ⁠ and slowly near $90^\circ$ ⁠, and never passes $1$ ⁠. So $\sin 60^\circ \approx 0.866$ ⁠, not $2 \cdot \sin 30^\circ = 1$ ⁠. Only for very small angles is it roughly true.
- Correct About $0.866$ ⁠: more than $0.5$ ⁠, less than $1$ ⁠
- Wrong $0.5$ ⁠: the same
  Feedback
  The bead climbs between $30^\circ$ ⁠ and $60^\circ$ ⁠, so its height grows.
- Wrong $0.25$ ⁠: half
  Feedback
  The bead climbs between $30^\circ$ ⁠ and $60^\circ$ ⁠, so its height grows, not shrinks.
Once the prediction is locked in, the Scene changes
Angle
60 °fixed
The Scene then shows: 0.866 · Unit-circle: sin θ, the bead's height/nodes/board/outputs/sin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ ⁠. The height grows, but not twofold: the nearer the bead gets to the top of the circle the slower it climbs, and at the top, at $90^\circ$ ⁠, it is only $1$ ⁠. So $\sin 2x \ne 2\sin x$ ⁠.

Worked solution
1. If $\sin 60^\circ$ ⁠ were $2 \cdot \sin 30^\circ = 1$ ⁠, the bead would be at the top of the circle already at $60^\circ$ ⁠. But it gets there only at $90^\circ$ ⁠.
2. At $60^\circ$ ⁠ the bead is partway from $30^\circ$ ⁠ to $90^\circ$ ⁠, between height $0.5$ ⁠ and height $1$ ⁠.
3. By Pythagoras: $\cos 60^\circ = 0.5$ ⁠, so $\sin 60^\circ = \sqrt{1 - 0.5^2} = \sqrt{0.75} \approx 0.866$ ⁠.
If you chose $1$ ⁠, you doubled the sine when you doubled the angle. The sine does not grow at a steady rate: $\sin 2x$ ⁠ is not $2\sin x$ ⁠.
Step 5 of 7
Open this step
Radians
Practice · Reach the target
Question
Set Angle as close as it can get to one radian, an arc as long as the radius, and press Check.

Goal
- 1 ± 0.01
  At the start of the step: 2.094
Setup, choices and feedback
The Scene at this step
Angle
120 °
Circle / Waves
Circle
Other settings
Angle unit/nodes/board/props/angleUnit = "radians"
Read from the Scene at
Unit-circle: the angle in radians, the arc it cuts from the unit circle/nodes/board/outputs/radians
One setting that reaches it
Angle
57 °
Feedback for a right answer
$57^\circ \approx 0.995$ ⁠ radians. One radian is $\frac{180^\circ}{\pi} \approx 57.3^\circ$ ⁠: an arc as long as the radius.

Feedback for a wrong answer
Not yet. The board writes the angle in radians beside the bead.

Hint
$90^\circ$ ⁠ is about $1.57$ ⁠ radians. One radian is a little more than half of that.

Worked solution
1. Half a turn is $\pi$ ⁠ radians, so $180^\circ = \pi$ ⁠ radians.
2. So one radian is $\frac{180^\circ}{\pi} \approx \frac{180^\circ}{3.14} \approx 57.3^\circ$ ⁠.
3. Angle moves in whole degrees: set it to $57^\circ$ ⁠, about $0.995$ ⁠ radians.
Step 6 of 7
Open this step
A full turn
Practice · Calculate
Question
Half a turn, $180^\circ$ ⁠, is $\pi \approx 3.14$ ⁠ radians. How many radians are there in a full turn, $360^\circ$ ⁠?

Correct answer
6.283 ± 0.01
Setup, choices and feedback
The Scene at this step
Angle
180 °fixed
Circle / Waves
Circle
Other settings
Angle unit/nodes/board/props/angleUnit = "radians"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Unit-circle: the angle in radians, the arc it cuts from the unit circle/nodes/board/outputs/radians
Wrong answers it recognises
- 3.142 ± 0.01
  Misconception: Thinks a full turn is π radians rather than 2π.
  Feedback from the catalogue
  $\pi$ ⁠ radians is half a turn, $180^\circ$ ⁠: the arc along half the unit circle. A full turn is the whole circumference, $2\pi \approx 6.28$ ⁠ radians.
- 360 ± 0.01
  Misconception: Ignores whether an angle (or a calculator) is in degrees or radians.
  Feedback from the catalogue
  The number $30$ ⁠ on its own does not say which unit it is in. $30^\circ$ ⁠ is $\frac{1}{12}$ ⁠ of a turn, but $30$ ⁠ radians is almost five full turns. A calculator in RAD mode reads every number as radians, and in DEG mode as degrees: check the mode before you trust the answer.
Feedback for a right answer
$2\pi \approx 6.28$ ⁠ radians: the circumference of the unit circle, $2\pi r$ ⁠ with $r = 1$ ⁠. Move Angle to $360^\circ$ ⁠ to see it.

Hint
A full turn is two half turns.

Worked solution
1. Radians measure the length of the arc on the unit circle.
2. A full turn is the whole circumference: $2\pi r = 2\pi \cdot 1 = 2\pi$ ⁠.
3. $2\pi \approx 2 \cdot 3.14 = 6.28$ ⁠ radians.
If you answered $3.14$ ⁠, that is half a turn. If you answered $360$ ⁠, that is the number in degrees, not in radians.
Step 7 of 7
Open this step
Degrees or radians on a calculator
Practice · Choose
Question
You type $\sin 30$ ⁠ into a calculator and get $-0.988$ ⁠, not $0.5$ ⁠. What happened?

Correct answer
- Correct The calculator is in radians, so it worked out the sine of $30$ ⁠ radians
Setup, choices and feedback
The Scene at this step
Angle
30 °
Circle / Waves
Circle
Choices
Shown in a random order.
- Correct The calculator is in radians, so it worked out the sine of $30$ ⁠ radians
- Wrong Nothing: $30$ ⁠ and $30^\circ$ ⁠ are the same angle, so $-0.988$ ⁠ is right
  Misconception: Ignores whether an angle (or a calculator) is in degrees or radians.
  Feedback from the catalogue
  The number $30$ ⁠ on its own does not say which unit it is in. $30^\circ$ ⁠ is $\frac{1}{12}$ ⁠ of a turn, but $30$ ⁠ radians is almost five full turns. A calculator in RAD mode reads every number as radians, and in DEG mode as degrees: check the mode before you trust the answer.
- Wrong The calculator rounded $0.5$ ⁠ wrongly
  Feedback
  Rounding does not flip a sign or turn $0.5$ ⁠ into $0.988$ ⁠. The calculator worked out the sine of a different angle.
Feedback for a right answer
In radians, $30$ ⁠ means $30$ ⁠ radians: four full turns and about $279^\circ$ ⁠ more, where the bead is very low, $\sin \approx -0.988$ ⁠. Switch the calculator to degrees, DEG, or convert: $30^\circ = \frac{\pi}{6} \approx 0.524$ ⁠ radians.

Hint
How many turns are $30$ ⁠ radians, if a full turn is about $6.28$ ⁠ radians?

Worked solution
1. A full turn is $2\pi \approx 6.28$ ⁠ radians, so $30$ ⁠ radians is $\frac{30}{6.28} \approx 4.77$ ⁠ turns.
2. Four full turns bring the bead back where it started, leaving $0.77$ ⁠ of a turn, about $279^\circ$ ⁠.
3. At $279^\circ$ ⁠ the bead is near the bottom of the circle: $\sin 279^\circ \approx -0.988$ ⁠. That is what the calculator showed.
4. To get $\sin 30^\circ = 0.5$ ⁠, switch to degrees.
If you chose that there is no difference, that is exactly the slip: a number without a unit means degrees or radians depending on the calculator’s mode.

Lesson 4 of 4 · About 25 minutes

Vectors and their components

A vector is an arrow: a length and a direction. Sine and cosine split it into a horizontal and a vertical component, and with the components you add vectors.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Components
Practice · Calculate
Question
The arrow $u$ ⁠ is $10$ ⁠ long, at $30^\circ$ ⁠ above the horizontal axis. What is its horizontal component, $u_x$ ⁠? ( $\cos 30^\circ \approx 0.866$ ⁠)

Correct answer
8.66 ± 0.05
Setup, choices and feedback
The Scene at this step
Length of u
10fixed
Direction of u
30 °fixed
Direction of v
90 °defaulthidden
Given / Worked out
Given
Other settings
Unknown/nodes/board/props/unknown = "vectorX"
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Vector: the horizontal component of u, length × cos direction/nodes/board/outputs/vectorX
Wrong answers it recognises
- 5 ± 0.05
  Misconception: Mixes up the opposite and the adjacent side, and so uses cos for sin or the reverse.
  Feedback from the catalogue
  The side opposite an angle does not touch it; the side adjacent to an angle is one of the two sides that make it. Sine uses the opposite side, cosine the adjacent one. Label all three sides before you pick a ratio.
- 0.866 ± 0.01
  Misconception: Gives the ratio itself as the answer, forgetting to multiply by the known side.
  Feedback from the catalogue
  $\sin A$ ⁠ is only the ratio, a number without units. To get a length, multiply it by the known side: if $\sin A = \frac{x}{c}$ ⁠, then $x = c \cdot \sin A$ ⁠.
Feedback for a right answer
$u_x = 10 \cdot \cos 30^\circ \approx 8.66$ ⁠. The arrow goes $8.66$ ⁠ to the right and rises $10 \cdot \sin 30^\circ = 5$ ⁠.

Hint
The horizontal component is the side adjacent to the angle, and the arrow is the hypotenuse.

Worked solution
1. The arrow, its horizontal component and its vertical component make a right triangle. The arrow is the hypotenuse, $10$ ⁠.
2. The horizontal component is the side adjacent to the $30^\circ$ ⁠ angle: adjacent and hypotenuse, cosine.
3. $u_x = 10 \cdot \cos 30^\circ \approx 10 \cdot 0.866 = 8.66$ ⁠.
If you got $5$ ⁠, you worked out the vertical component, with the sine. If you got $0.866$ ⁠, you forgot to multiply by the length.
Step 2 of 6
Open this step
A negative component
Practice · Reach the target
Question
Length of u is locked at $6$ ⁠. Turn Direction of u so that its horizontal component is $-3$ ⁠, and press Check.

Goal
- -3 ± 0.05
  At the start of the step: 5.196
Setup, choices and feedback
The Scene at this step
Length of u
6fixed
Direction of u
30 °
Direction of v
90 °defaulthidden
Given / Worked out
Worked outfixed
Read from the Scene at
Vector: the horizontal component of u, length × cos direction/nodes/board/outputs/vectorX
One setting that reaches it
Direction of u
120 °
Feedback for a right answer
$6 \cdot \cos 120^\circ = 6 \cdot (-0.5) = -3$ ⁠. $240^\circ$ ⁠ works too: there the arrow points left and down, and the vertical component is negative as well.

Feedback for a wrong answer
Not yet. The horizontal component is negative only when the arrow points left, between $90^\circ$ ⁠ and $270^\circ$ ⁠.

Hint
You need $6 \cdot \cos\theta = -3$ ⁠, that is, $\cos\theta = -0.5$ ⁠. Where on the unit circle is the width $-0.5$ ⁠?

Worked solution
1. $u_x = 6 \cdot \cos\theta$ ⁠, and it has to be $-3$ ⁠.
2. So $\cos\theta = \frac{-3}{6} = -0.5$ ⁠.
3. $\cos 60^\circ = 0.5$ ⁠. The cosine is $-0.5$ ⁠ at the mirror image across the vertical axis: $180^\circ - 60^\circ = 120^\circ$ ⁠ (and at $240^\circ$ ⁠).
4. Turn Direction of u to $120^\circ$ ⁠.
Step 3 of 6
Open this step
From components to length
Practice · Calculate
Question
The vector $u$ ⁠ has horizontal component $3$ ⁠ and vertical component $4$ ⁠. How long is it?

Correct answer
5 ± 0.02
Setup, choices and feedback
The Scene at this step
Length of u
5hidden
Direction of u
53.13 °hidden
Direction of v
90 °defaulthidden
Given / Worked out
Given
Other settings
Unknown/nodes/board/props/unknown = "vectorLength"
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Vector: the length of u/nodes/board/outputs/vectorLength
Wrong answers it recognises
- 7 ± 0.02
  Misconception: Adds vector lengths (or a vector's components) as plain numbers, ignoring directions.
  Feedback from the catalogue
  Lengths add directly only when the arrows point the same way. Otherwise add component by component: the horizontal components together, and the vertical ones together. Find the length of the result with Pythagoras’ theorem; it is never more than the sum of the lengths.
- 25 ± 0.02
  Misconception: Adds vector lengths (or a vector's components) as plain numbers, ignoring directions.
  Feedback
  $25$ ⁠ is $3^2 + 4^2$ ⁠, the square of the length. You still need the square root.
Feedback for a right answer
$|u| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ ⁠: Pythagoras’ theorem. The components do not add like plain numbers, because they point in different directions.

Hint
The components are the legs of a right triangle, and the arrow is its hypotenuse.

Worked solution
1. The horizontal component, $3$ ⁠, and the vertical component, $4$ ⁠, are at right angles. The arrow is the hypotenuse of the triangle they make.
2. Pythagoras: $|u|^2 = 3^2 + 4^2 = 9 + 16 = 25$ ⁠.
3. $|u| = \sqrt{25} = 5$ ⁠.
If you answered $7$ ⁠, you added the components. That is the length of the walk if you go right first and then up, not the straight-line distance.
Step 4 of 6
Open this step
Adding vectors
Practice · Predict, then look
Question
You walk $4$ ⁠ steps east, then $3$ ⁠ steps north. How far are you, in a straight line, from where you started?

Correct answer
- Correct $5$ ⁠
Setup, choices and feedback
The Scene at this step
Length of u
4fixed
Direction of u
0 °fixed
Direction of v
90 °fixed
Given / Worked out
Givenfixed
Other settings
Vector2 length/nodes/board/props/vector2Length = 3
Other settings
Unknown/nodes/board/props/unknown = "sumLength"
Choices
- Correct $5$ ⁠
- Wrong $7$ ⁠
  Misconception: Adds vector lengths (or a vector's components) as plain numbers, ignoring directions.
  Feedback from the catalogue
  Lengths add directly only when the arrows point the same way. Otherwise add component by component: the horizontal components together, and the vertical ones together. Find the length of the result with Pythagoras’ theorem; it is never more than the sum of the lengths.
- Wrong $1$ ⁠
  Feedback
  $4 - 3 = 1$ ⁠ is right only when you walk back the way you came. North is sideways.
- Wrong $12$ ⁠
  Feedback
  $4 \cdot 3 = 12$ ⁠ is the area of a rectangle, not a distance.
Once the prediction is locked in, the Scene changes
Given / Worked out
Worked outfixed
The Scene then shows: 5 · Vector: the length of u + v, by Pythagoras from its components/nodes/board/outputs/sumLength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$u + v = (4, 3)$ ⁠, and its length is $\sqrt{4^2 + 3^2} = 5$ ⁠. You walked $7$ ⁠ steps, but you are only $5$ ⁠ from the start: lengths add directly only when you keep walking the same way.

Worked solution
1. $u = (4, 0)$ ⁠ and $v = (0, 3)$ ⁠. Add component by component: $u + v = (4 + 0, 0 + 3) = (4, 3)$ ⁠.
2. The sum is the hypotenuse of a right triangle with legs $4$ ⁠ and $3$ ⁠.
3. $|u + v| = \sqrt{16 + 9} = \sqrt{25} = 5$ ⁠.
If you chose $7$ ⁠, you added the lengths. $7$ ⁠ is how far you walked, not how far you are from the start.
Step 5 of 6
Open this step
Component by component
Practice · Calculate
Question
$u = (6, 2)$ ⁠ and $v = (-2, 3)$ ⁠. How long is $u + v$ ⁠?

Correct answer
6.403 ± 0.02
Setup, choices and feedback
The Scene at this step
Length of u
6.325hidden
Direction of u
18.43 °hidden
Direction of v
123.7 °hidden
Given / Worked out
Given
Other settings
Vector2 length/nodes/board/props/vector2Length = 3.605551275
Other settings
Unknown/nodes/board/props/unknown = "sumLength"
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Vector: the length of u + v, by Pythagoras from its components/nodes/board/outputs/sumLength
Wrong answers it recognises
- 9.93 ± 0.03
  Misconception: Adds vector lengths (or a vector's components) as plain numbers, ignoring directions.
  Feedback from the catalogue
  Lengths add directly only when the arrows point the same way. Otherwise add component by component: the horizontal components together, and the vertical ones together. Find the length of the result with Pythagoras’ theorem; it is never more than the sum of the lengths.
- 41 ± 0.02
  Misconception: Adds vector lengths (or a vector's components) as plain numbers, ignoring directions.
  Feedback
  $41$ ⁠ is $4^2 + 5^2$ ⁠, the square of the length. You still need the square root.
Feedback for a right answer
$u + v = (4, 5)$ ⁠, and $\sqrt{4^2 + 5^2} = \sqrt{41} \approx 6.40$ ⁠. The sum of the lengths, $6.32 + 3.61 \approx 9.93$ ⁠, is much longer: the horizontal components pull in opposite directions.

Hint
Add the components first: $6 + (-2)$ ⁠ and $2 + 3$ ⁠. Then Pythagoras.

Worked solution
1. Horizontal: $6 + (-2) = 4$ ⁠. Vertical: $2 + 3 = 5$ ⁠. So $u + v = (4, 5)$ ⁠.
2. Pythagoras: $|u + v|^2 = 4^2 + 5^2 = 16 + 25 = 41$ ⁠.
3. $|u + v| = \sqrt{41} \approx 6.40$ ⁠.
If you got about $9.93$ ⁠, you added the lengths of $u$ ⁠ and $v$ ⁠. That is right only when they point the same way.
Step 6 of 6
Open this step
Vectors at right angles
Practice · Reach the target
Question
$u$ ⁠ is $4$ ⁠ long, pointing east, and $v$ ⁠ is $3$ ⁠ long. Turn Direction of v so that $u + v$ ⁠ is exactly $5$ ⁠ long, and press Check.

Goal
- 5 ± 0.02
  At the start of the step: 7
Setup, choices and feedback
The Scene at this step
Length of u
4fixed
Direction of u
0 °fixed
Direction of v
0 °
Given / Worked out
Worked outfixed
Other settings
Vector2 length/nodes/board/props/vector2Length = 3
Read from the Scene at
Vector: the length of u + v, by Pythagoras from its components/nodes/board/outputs/sumLength
One setting that reaches it
Direction of v
90 °
Feedback for a right answer
At $90^\circ$ ⁠ (or $270^\circ$ ⁠) $v$ ⁠ is at right angles to $u$ ⁠, and the sum is the hypotenuse: $\sqrt{4^2 + 3^2} = 5$ ⁠. In other directions the length ranges from $4 - 3 = 1$ ⁠ to $4 + 3 = 7$ ⁠.

Feedback for a wrong answer
Not yet. When the directions are the same the length is $7$ ⁠, and when they are opposite, $1$ ⁠. Where in between?

Hint
$5$ ⁠ is the hypotenuse of a triangle with legs $4$ ⁠ and $3$ ⁠. What angle must there be between $u$ ⁠ and $v$ ⁠?

Worked solution
1. The lengths $4$ ⁠, $3$ ⁠ and $5$ ⁠ are the triangle from the geometry Unit: $3^2 + 4^2 = 5^2$ ⁠.
2. So $u + v$ ⁠ is $5$ ⁠ long when $u$ ⁠ and $v$ ⁠ are at right angles.
3. $u$ ⁠ points at $0^\circ$ ⁠, so $v$ ⁠ has to point at $90^\circ$ ⁠ or $270^\circ$ ⁠.
4. Turn Direction of v to $90^\circ$ ⁠.

Across the unit · 13 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Sine, cosine and tangent

Naming the sides

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Same angle, same ratio

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Working out a sine

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Cosine

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Tangent

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Finding sides and angles

A ladder against a wall

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A ramp

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The height of a tree

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

From a ratio to an angle

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The inverse function

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The unit circle and radians

More than 90 degrees

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution