Trigonometry (20-1)

Question 1

Solve for x:
sin(x) = 1/2, 0° ≤ x ≤ 360°

Answer 1

Access sin^-1 then (1÷2)

x = 30°, 150°

The calculator only returns 30 degrees, but you must also look for the other quadrant where sine is positive, and then make sure to satisfy the restrictions given.

This can also be memorized by using the 30-60 degree triangle. Since sine is opposite over hypotenuse, you’ll know that it is equal to 30 degrees. Even so, you need to use that as the reference angle and recognize where sine is positive (quadrant I and quadrant II). 180 degrees - 30 degrees will return the 150 degrees.

Question 2

Solve for x:
cos(x) = 0, 0° ≤ x ≤ 360°

Answer 2

x = 90°, 270°

I use the graph that I have memorized with period 360 degrees to understand this one. For cosine: Up, through, down, through is the wave, and it takes 360 degrees to complete the wave. 90 degrees is the first through, and 270 degrees is three quarters of the way through the period.

Your calculator will give you 90 degrees, but you have to figure out the 270 degrees.

You can also use triangles in each of the quadrants and know that cosine is adjacent (x on the graph) over hypotenuse (1 on the unit circle), so cos(theta) = x. Then we look at where x is equal to zero, and that is at 90 degrees and 270 degrees.

Question 3

Find the missing side:
right triangle with hypotenuse 13, one leg 5.

Answer 3

Other leg = 12

This is Pythagorean Theorem review.

13 squared - 5 squared, then square root that answer

This is a Pythagorean triple and worth memorizing because it comes up more often than others.

Question 4

Find the missing angle:
right triangle with opposite = 4, hypotenuse = 5. Round the angle to the nearest 100th.

Answer 4

Use Sine since we have a right triangle with opposite and hypotenuse given.

sin(θ) = opp / hyp
sin(θ) = 4 / 5
θ = sin^-1 (4/5)
θ ≈ 53.13°

This should be review from a previous grade (old grade 10)

Notice that you must use a calculator here because the ratio is not one of the ones that you have to memorize.

Question 5

Solve for side c:
a=7, b=9, C=120° (use cosine law).

Answer 5

c² =7²+9²−2(7)(9)cos(120∘)
c² =49+81−126cos(120∘)
c² =49+81−126(-1/2)
c² =49+81+63
c² = 193
c ≈ 13.89

This is not the ambiguous case because we have the contained angle

In other questions watch out for the ambiguous case with SSA. This is SAS, so we do not need to worry about it.

Question 6

Solve for angle A:
a=8, b=10, c=12 (use cosine law).

Answer 6

The rearranged formula is probably present on your formula sheet, but you can rearrange it yourself using the other cosine law formula too. I’ve showed the already rearranged formula here.

cos A = (100 + 144 - 64) / (2(10)(12))
cos A = 180 / 240
cos A = 3/4

A ≈ 41.4°

Question 7

Solve for angle A:
a=7, b=10, c=12 (use cosine law).

Answer 7

a² = b² + c² -2bcCosA

7² = 10² + 12² -2(10)(12)CosA

-195/(-240) = CosA

13/16 = CosA

A ≈ 36.0°

Question 8

Determine if there are 0, 1, or 2 solutions: a=8, b=10, A=40°.

Answer 8

2 solutions (ambiguous SSA case)

Set up Sine Law:
SinA / a = SinB / b
Sin40 / 8 = SinB / 10
0.8 = SinB
sin-1(0.8) = B
53.46 degrees, 126.54 degrees = B

There are two angles possible. We must check if both triangles are actually possible though.

We will check if Sin is valid. Since it is 0.8 = SinB and 0.8 is equal to or between 0 and 1, we know that Sine is valid. This means there is at least one solution to the triangle.

Use both cases to see if the angle is possible:
1. Case 1 is that angle B is 53.46 degrees while Angle A is 40 degrees as given. Is it possible to have an angle C using the angle sum rule of 180 degrees? 180 - 40 - 53.46 = a positive number so this is possible. (Angle C could be 86.54 degrees here)
2. Case two is that angle B is 126.54 degrees while Angle A is 40 degrees as given. Use the angle sum to make sure a third angle is actually possible. 180 - 40 - 126.54 = a positive number so this is possible (Angle C could be 13.46 degrees).

Therefore there are two solutions.

If SinB = 1 then there would have been a right triangle with exactly one solution. That was not the case here, but I wrote this here for reference with future questions.

In the SSA ambiguous case, you always begin by finding the sine of angle B using the Law of Sines. If the sine of angle B is greater than 1 (or negative), no triangle exists. If the sine of angle B equals 1, then angle B is 90° and there is exactly one solution, a right triangle. If the sine of angle B is less than 1, first find the acute angle B. Then check the second possible angle, which is 180° minus B. If angle A plus this second possible angle is greater than or equal to 180°, there is only one triangle (the obtuse option does not work). If angle A plus this second possible angle is less than 180°, there are two possible triangles: one with angle B acute and one with angle B obtuse.

Question 9

Find exact value:
tan(225°)

You should be able to do this without a calculator.

Answer 9

The reference angle is 225 -180 = 45 degrees

Remember that the reference angle is the measurement of the angle as the closest way to get to that angle from the x axis (positive or negative x-axis can be used) so the reference angle should be under 90 degrees inclusive.

tan(225°) = tan(45°)
= 1/1 because tan is opposite over adjacent (see picture of memorized triangle below)
= 1

Memorize the 45 degree triangle to get this one.

Question 10

Find exact value:
cos(330°)

You should be able to do this without a calculator.

Answer 10

The reference angle is 360 degrees - 330 degrees = 30 degrees

Remember that the reference angle is the closest way of measuring that angle to the x-axis (think of a bird flapping its wings!)

cos(330°) = cos(30°)
= √3/2 because cosine is adjacent divided by hypotenuse in the memorized triangle as pictured below

Question 11

State amplitude and period:
y = 3sin(2x)

Answer 11

Amplitude = 3, Period = 180°

The picture below is the sine graph before transforming it. After transformation, it went through a vertical stretch by a factor of 3 and a horizontal compression making it half its width.

The sine graph normally has an amplitude of 1 (distance from the midline to the extrema), and a period of 360 degrees (period is how long it takes for the pattern to repeat itself). That amplitude is multiplied by 3 because a = 3, and the period is divided by 2 because the reciprocal of 2 is 1/2, and we must multiply by the reciprocal to get the horizontal stretch.

Question 12

Find x-intercepts of y = cos(x) on 0° ≤ x ≤ 360°.

Answer 12

x = 90°, 270°

If you memorize everything in the picture below, a lot of these types of questions are much easier. Memorize all three graphs (sine, cosine, and tangent graphs). Learn the period of each (sine and cosine are 360 degrees, and tan is 180 degrees), as well as the extrema for sine and cosine (lowest value is -1 and highest value is 1).

Question 13

What is the checklist for the ambiguous case?

Answer 13

SSA Ambiguous Case Checklist

Find the sine of angle B using the Law of Sines.

1. Check if the sine is valid:

If the sine of angle B is greater than 1 (or negative), there is no solution.

2. If the sine of angle B equals 1, angle B is 90° and there is one solution (a right triangle).
3. If the sine is less than 1:

Find the acute angle B using the inverse sine function.

Find the second possible angle by subtracting B from 180°.

Add angle A to this second possible angle.

If the sum is greater than or equal to 180°, there is only one triangle (use the acute angle B).

If the sum is less than 180°, there are two possible triangles (one with angle B acute and one with angle B obtuse).

Question 14

What are coterminal angles? Describe in terms of degrees and in terms of radians.

Answer 14

Coterminal angles are angles that share the same terminal side. They differ by a multiple of 360° (or 2π radians).

So these angles look like they are at the same position in the quadrant, but one of them may have gotten there by doing an extra rotation or rotated backwards to get there. Think of the hands on a clock that moves counterclockwise. Every rotation produces another name for the same position.

Add 360 degrees repeatedly to find coterminal angles that are represented with a number greater than the given angle. Subtract 360 degrees repeatedly to find coterminal angles that are represented with a number less than the given angle. You can do something similar with radians, but use 2 pi instead of 360 degrees.

Question 15

What is a reference angle?

Answer 15

A reference angle is the acute angle (angle under 90 degrees) formed between the terminal side of an angle and the x-axis. It helps determine trig values in all quadrants because the calculator only returns one answer, but you need to find the other angle within 360 degrees using the reference angle. For more angles, use the idea of coterminal angles as described on the coterminal angle card.

Question 16

What are the three basic trigonometry ratios for a right triangle?

Answer 16

sin(θ) = opposite / hypotenuse

cos(θ) = adjacent / hypotenuse

tan(θ) = opposite / adjacent

SOH CAH TOA

Where theta (θ) is the angle, adjacent is the side touching the angle but not the longest side of the triangle, and opposite is the side not touching the angle of concern. Hypotenuse is the longest side of the triangle. The other two sides are called legs.

Question 17

What are the reciprocal trigonometric functions? Relate them to opposite, adjacent, and hypotenuse

Answer 17

cotangent:
cot(θ) = 1 / tan(θ)
= adjacent / opposite

secant:
sec(θ) = 1 / cos(θ)
= hypotenuse / adjacent

cosecant:
csc(θ) = 1 / sin(θ)
= hypotenuse / opposite

Question 18

How do you find a coterminal angle? Which should you report when giving your answer?

Answer 18

To find a coterminal angle, add or subtract 360° (or 2π radians) until the angle is in the desired range (usually the desired range is implied to be 0° to 360° or 0 to 2π for radians when you are first learning).

If no range is given, at the higher levels they expect a general answer using n, such as 21° + 360°n to show all angles coterminal to 21°, where n is an integer. Similarly you could report in radians using 2 pi.

Question 19

What is the CAST rule, and why is it useful?

Answer 19

When working with trig in different quadrants:
- sine is positive in QI & QII
- cosine is positive in QI & QIV
- tangent is positive in in QI & QIII

See the picture below for a visual. Memorizing the graphs can also allow you to see when the sine of an angle is positive, and you can see that in the graph showing that the curve is above the x axis for the first half, so between 0 degrees and 180 degrees, which corresponds to quadrant 1 and quadrant 2 when looking at the rotation of the angle counterclockwise about the origin.

Similar reasoning can be applied to the other two basic trigonometry ratios using those memorized graphs. You can also use the same signs for the reciprocals since they include the same two numbers in their ratios as the basic ratios had.

We use the CAST rule to remember which quadrant to choose when reporting the second angle within 360 degrees. The reference angle gets us four angles, but we use the sign (positive or negative) to select the correct two angles.

Question 20

Why should you check for the ambiguous case before using the Law of Sines to find a missing angle?

Example: A = 40°, a = 8, b = 10, find B.

Answer 20

Because in SSA, there may be 0, 1, or 2 possible triangles. Using sin⁻¹ blindly may give only 1 solution, causing you to miss another possible triangle.

In the picture below, you will see the example worked out. While the diagrams are not to scale, you can see that two triangles are possible, and that you must find them by realizing the sine is positive in both quadrant I and quadrant II and thus you need to check if the quadrant II angle might fit the situation as well. The quadrant I angle will be reported back by your calculator, but you need to calculate the quadrant II angle and then see if it is possible using the angle sum of interior angles in a triangle being 180 degrees. If the two angles (given, and one of the calculated angles) add to less than 180 degrees, then you have found a possible angle for B. In the case pictured below, that is exactly what happens, two angles come out of sine using the CAST rule and then we notice that both work. So two triangles are possible.

Example: A = 40°, a = 8, b = 10. Using sin⁻¹ directly for angle B gives only one angle, but there is actually a second possible triangle.

Question 21

What is the tip for determining the number of triangles before using the Law of Sines?

Explain with the following example:
A = 40 degrees, b = 10, a = 8

Answer 21

Always compute h = b × sin(A) first to see how many triangles exist before finding missing angles with Law of Sines. bSinA is less the other side length given in our example on the previous card that contained two triangles. (A = 40 degrees, b = 10, a = 8)

h = bsinA
= 10sin(40 degrees)
= 6.427876

Since h < a < b, 2 triangles exist → ambiguous case must be considered.

Explanation: compare that height to the other side lengths given and it should be shorter than them for the two triangles to be possible. Generally, if you miss this step you can still continue and watch out for sine theta values being within range.

I actually do not bother calculating the height, but instead compute sine to determine an angle and report both angles back using the CAST rule. Then I check if both fit inside a 180 degree interior sum triangle. Sometimes when computing sine you will notice an error on your calculator due to sine being equal to a number greater than 1 or less than -1, and that means that no triangles exist. You already have memorized your sine graph and know that the sine ratio must be between -1 and 1 inclusive. So values outside of this should signal that no triangle exists. If sine is equal to 1, then we have exactly one triangle with 90 degrees.

Question 22

Practice: A = 35°, a = 6, b = 8. Determine if an ambiguous case exists before using Law of Sines for a missing angle.

Answer 22

h = b × sin(A) = 8 × sin35° ≈ 4.59. Since h < a < b, 2 triangles exist → ambiguous case must be considered.

Question 23

How do you determine the number of possible triangles in an SSA (Ambiguous Case) problem?

Answer 23

In SSA, you are given: one angle (A), its opposite side (a), and another side (b).
Step 1: Compute the height h = b × sin(A)
Step 2: Compare a to h and b:
- If a < h → 0 triangles
- If a = h → 1 triangle (right triangle)
- If h < a < b → 2 triangles
- If a ≥ b → 1 triangle

Question 24

SSA Practice 1: A = 40°, a = 6, b = 10. How many triangles exist?

Answer 24

h = 10 × sin40° ≈ 6.43 → a < h → 0 triangles exist.

Question 25

SSA Practice 2: A = 50°, a = 8, b = 10. How many triangles exist?

Answer 25

h = 10 × sin50° ≈ 7.66 → a > h and a < b → 2 triangles exist.

Question 26

SSA Practice 3: A = 60°, a = 9, b = 8. How many triangles exist?

Answer 26

h = 8 × sin60° ≈ 6.93 The height is smaller than a and smaller than b. a > b → 1 triangle exists.

Question 27

SSA Practice 4: A = 35°, a = 4, b = 7. How many triangles exist?

Answer 27

h = 7 × sin35° = 4.02 → a is smaller than h → 0 triangles exist.

Question 28

SSA Practice 5: A = 45°, a = 3, b = 5. How many triangles exist?

Answer 28

h = 5 × sin45° ≈ 3.54 → a < h → 0 triangles exist.

Question 29

SSA Practice 6: A = 20°, a = 7, b = 10. How many triangles exist?

Answer 29

h = 10 × sin20° ≈ 3.42 → a > h and a < b → 2 triangles exist.

Question 30

What is the ambiguous case in the Law of Sines (SSA)?

Answer 30

The ambiguous case occurs when two sides and a non-included angle (SSA) are given, and there may be 0, 1, or 2 possible triangles. ## Footnote Example: Given angle A = 40°, side a = 8, side b = 10. How many triangles exist? Answer: There are 2 possible triangles because h < a < b. h = bsinA = 6.4

Question 31

How do you find a missing side using the Law of Sines?

Answer 31

Use the formula: sin A / a = sin B / b = sin C / c. Solve for the unknown. ## Footnote Example: A = 40°, a = 8, B = 70°, find b. b = (sin 70° / sin 40°) × 8 ≈ 13.2.

Question 32

How do you find a missing angle using the Law of Sines?

Answer 32

sin B = (b sin A) / a. Then use arcsin to find B. ## Footnote Example: A = 40°, a = 8, b = 10, find B. sin B = (10 × sin 40°)/8 ≈ 0.856 → B ≈ 59°.

Question 33

How do you solve a triangle using the Law of Cosines for a missing side?

Answer 33

Use c² = a² + b² − 2ab cos C. Solve for c. ## Footnote Example: a = 7, b = 10, C = 60°, find c. c² = 7² + 10² − 2(7)(10) cos 60° → c² = 49 + 100 − 70 → c² = 79 → c ≈ 8.89.

Question 34

How do you solve a triangle using the Law of Cosines for a missing angle?

Answer 34

Use cos C = (a² + b² − c²)/(2ab). Solve for C with arccos. ## Footnote Example: a = 7, b = 10, c = 8.89, find C. cos C = (49 + 100 − 79) / (2×7×10) = 70/140 = 0.5 → C = 60°.

Question 35

How do you find the area of a non-right triangle using trigonometry?

Answer 35

Use Area = (½)cb sin A, where c and b are sides, and A is the included angle. This is because triangle area is 0.5hb, and bsinC gives a height at the vertex C, c is the base ## Footnote Example: c = 7, b = 10, A = 60°, Area = ½ × 7 × 10 × sin 60° ≈ 30.3.

Question 36

Sine Graph

Answer 36

Question 37

Cosine Graph

Answer 37

Question 38

Tangent Graph

Answer 38