Finding missing angles using inverse trig — Mathematics

Before this

This leads to

Angles of elevation and depression
The sine rule
Exact values and the unit circle · coming soon

A safety rule that gives two sides and hides the angle

Safety guidance for leaning ladders gives a simple rule. Put the foot of the ladder 1 unit out from the wall for every 4 units up. The same guidance calls this a 75° angle with the ground. (Source: UK Health and Safety Executive, Safe use of ladders and stepladders, INDG455.)

The "1 out for every 4 up" rule. The two shorter sides are known. The angle θ between the ladder and the ground is not.

Is that true? Is a ladder that is 1 out and 4 up really at 75°?

Look at what you know. From the angle $θ$ at the foot, the height up the wall is the opposite side, 4 units. The distance out is the adjacent side, 1 unit. Opposite and adjacent means tangent, so you can write the equation:

tan θ = \frac{4}{1} = 4

In the last lecture, every equation had a missing side. You found it by multiplying or dividing. Here both sides are known, and the unknown is the angle, inside the tangent. Dividing both sides by "tan" makes no sense, because tan is not a number. It is an instruction: "take this angle and give me its ratio". So the question becomes a new one. Which angle has a tangent of 4? That question is what this lecture answers.

Searching for the angle by hand

You can already answer it slowly, with the calculator you have. Guess an angle, find its tangent, and adjust.

Guess for $θ$	$tan θ$	Too big or too small?
70°	2.7475	too small: try a bigger angle
75°	3.7321	still too small
76°	4.0108	a little too big
75.9°	3.9812	a little too small

So the angle is between 75.9° and 76°, very close to 76°. The search works, and it shows what the answer means: an angle whose tangent is 4. It also settles the ladder question. At exactly 75°, the tangent is only about 3.73, not 4. The "1 in 4" rule gives an angle of about 76°, and the guidance rounds it to the friendly number 75°. The difference is about 1°, which does not matter for a ladder. But you now know which of the two is exact.

Why did the search have exactly one answer? You saw the reason in the first trigonometry lecture. As the angle grows from 0° to 90°, the opposite side grows and the adjacent side shrinks, so the tangent rises steadily. It never comes back down. The same is true of the sine, while the cosine falls steadily. So each ratio belongs to exactly one angle between 0° and 90°. A function that never repeats an output can be run backwards. That is the key.

Running the function backwards

Think of sine as a machine. You put in an angle, and it gives out a ratio. For example, $sin 3 0^{\circ} = 0.5$ . In the lecture on inverse functions, you met the machine that undoes a function. It takes the output and gives back the input. The inverse of sine takes the ratio 0.5 and gives back the angle, 30°.

Sine (solid arrows) takes an angle to its ratio. Inverse sine, written sin⁻¹ (dashed arrows), takes the ratio back to the angle. Every arrow pair has the same two labels as the top row.

Inverse sine, cosine and tangent

$sin^{- 1} (x)$ is the angle whose sine is $x$ . $cos^{- 1} (x)$ is the angle whose cosine is $x$ . $tan^{- 1} (x)$ is the angle whose tangent is $x$ .

So $sin θ = x$ means exactly the same as $θ = sin^{- 1} (x)$ , for an angle $θ$ in a right-angled triangle.

Read $sin^{- 1} (x)$ as "inverse sine of $x$ ". The $- 1$ is part of the function's name, exactly as in $f^{- 1}$ . It does not mean "one over". This matters, and Trap 1 returns to it.

Now the ladder takes one line. On most calculators the inverse is the second job of the same key, so you press SHIFT (or 2nd), then tan, then 4:

θ = tan^{- 1} (4) \approx 75.9 6^{\circ} \approx 76. 0^{\circ}

To 1 decimal place, that is 76.0°: the same answer as the search, found in one step. The calculator does the search for you, very fast and very precisely.

That is it. That is the whole idea. The ratio is an output of sine, cosine or tangent. The inverse function takes the output back to the angle.

Before you trust the calculator, test its mode. Work out $sin^{- 1} (0.5)$ . You should get 30. If you get about 0.524, the calculator is in radian mode, which measures angles in a different unit. Change it to degree mode.

Same ratio, same angle

An angle depends only on the ratio of two sides, not on their sizes. Try this below. Change the two sides one at a time and watch the angle.

Explore — the ratio decides the angle

Set the opposite side O and the hypotenuse H. The angle θ comes from sin⁻¹(O ÷ H). Try 3 and 6, then 5 and 10. Then make O longer than H.

Try:

Opposite O: 6 cm

Hypotenuse H: 10 cm

This table shows several pairs of sides, the ratio O ÷ H (the sine of θ), and the angle θ = sin⁻¹(O ÷ H), to 1 decimal place.

Opposite O (cm)	Hypotenuse H (cm)	O ÷ H	Angle θ
3	6	0.5	30.0°
5	10	0.5	30.0°
6	10	0.6	36.9°
8	10	0.8	53.1°
9	10	0.9	64.2°
12	10	1.2	no angle: no triangle can be drawn

Different sides with the same ratio give the same angle. A ratio above 1 gives no angle, because the opposite side cannot be longer than the hypotenuse.

You should have seen two things. First, 3 and 6 give exactly the same angle as 5 and 10, 30°, because the ratio is 0.5 both times. This is the similar-triangles idea from the first trigonometry lecture, now running backwards. Second, when the opposite side is longer than the hypotenuse, there is no triangle at all.

Stop and think

A student works out $sin^{- 1} (1.2)$ , and the calculator shows an error. Is the calculator broken?

Show the answer

No. The calculator is right. The sine of an angle in a right-angled triangle is opposite ÷ hypotenuse. The hypotenuse is the longest side, so this ratio is always less than 1. No angle has a sine of 1.2, so there is no answer to give. In the language of the inverse functions lecture, the inputs that $sin^{- 1}$ accepts are the outputs that sine can give. The same is true for cosine. If you ever get this error, check your working. You have almost certainly put the sides the wrong way up, such as hypotenuse ÷ opposite. Tangent is different: opposite ÷ adjacent can be any positive number, so $tan^{- 1} (4)$ works.

The method, with a reason for each step

Choosing the inverse works exactly like choosing the ratio. Only the last step is new.

1. Mark the angle you want and name the sides. Name the hypotenuse first, across from the right angle. Then name the opposite and the adjacent from the unknown angle. This is because the names only make sense from that angle.
2. Find the pair of known sides. Opposite and hypotenuse means sine. Adjacent and hypotenuse means cosine. Opposite and adjacent means tangent.
3. Write the equation. For example, $sin θ = \frac{8}{13}$ . Writing it shows that the unknown is the angle, not a side.
4. Apply the matching inverse. $θ = sin^{- 1} (\frac{8}{13})$ . Type the whole fraction into the calculator, so no rounding happens before the inverse.
5. Round and check. Give the angle to 1 decimal place (1 d.p.). Then check it, using one of the tests below.

Three checks are quick. First, compare the two shorter sides. If the opposite is shorter than the adjacent, the angle is below 45°. If it is longer, the angle is above 45°. Second, find the other angle with the other ratio. The two must add up to 90°. Third, put the angle back in: the sine of your answer should give back your ratio.

Here is a question about how the ratio and the angle are connected. Commit to an answer before you open the explanation.

Predict — double the tangent, double the angle?

You know that $tan 4 5^{\circ} = 1$ . Now $tan θ = 2$ . What is $θ$ ? Choose one, then read the explanation.

This box compares the angles whose tangents are 1 and 2.

$tan^{- 1} (1) = 4 5^{\circ}$ , but $tan^{- 1} (2) \approx 63. 4^{\circ}$ , not 90°. Doubling the ratio does not double the angle. At 90°, the triangle would be flat, and the adjacent side would be zero, so 90° has no tangent at all.

The same is true of sine. $sin 3 0^{\circ} = 0.5$ , but the angle with a sine of 0.25 is about 14.5°, not 15°. The ratio and the angle are linked by a curve, not a straight line. This is why there is no shortcut by proportion, and why the inverse button exists.

Worked examples

Worked Example 1 — Opposite and hypotenuse

In a right-angled triangle, the side opposite angle $θ$ is 8 cm, and the hypotenuse is 13 cm. Find $θ$ .

Think first. The pair is opposite and hypotenuse, so the ratio is sine.

sin θ = \frac{8}{13}

θ = sin^{- 1} (\frac{8}{13}) \approx 37.9 8^{\circ} \approx 38. 0^{\circ}

Check. By Pythagoras, the third side is $1 3^{2} - 8^{2} = 105 \approx 10.25$ cm. The opposite (8 cm) is shorter than the adjacent (10.25 cm), so the angle should be below 45°. It is. ✓

Worked Example 2 — Adjacent and hypotenuse

A right-angled triangle has a hypotenuse of 11 cm. The side next to angle $θ$ , which is not the hypotenuse, is 7 cm. Find $θ$ , and then the third angle.

Think first. "Next to $θ$ and not the hypotenuse" is the adjacent. Adjacent and hypotenuse means cosine.

cos θ = \frac{7}{11}

θ = cos^{- 1} (\frac{7}{11}) \approx 50.4 8^{\circ} \approx 50. 5^{\circ}

The third angle is $9 0^{\circ} - 50.4 8^{\circ} \approx 39. 5^{\circ}$ .

Check. From the third angle, the 7 cm side is now the opposite. So $sin^{- 1} (\frac{7}{11})$ should also give the third angle. It gives about 39.52°, so 39.5°. ✓

Worked Example 3 — Opposite and adjacent

A right-angled triangle has an opposite side of 9 cm and an adjacent side of 4 cm, measured from $θ$ . Find $θ$ .

Think first. There is no hypotenuse in the data. Opposite and adjacent means tangent.

tan θ = \frac{9}{4} = 2.25

θ = tan^{- 1} (2.25) \approx 66.0 4^{\circ} \approx 66. 0^{\circ}

Check. The opposite is longer than the adjacent, so the angle should be above 45°. It is. ✓ The other angle is $tan^{- 1} (\frac{4}{9}) \approx 23.9 6^{\circ}$ , and $66.0 4^{\circ} + 23.9 6^{\circ} = 9 0^{\circ}$ . ✓

Worked Example 4 — Both angles, and why sine and cosine swap

A right-angled triangle has sides of 6 cm, 8 cm and 10 cm. Find both of its acute angles.

Think first. The hypotenuse is the longest side, 10 cm. Call the angle opposite the 6 cm side $α$ , and the angle opposite the 8 cm side $β$ .

$sin α = \frac{6}{10} = 0.6$ , so $α = sin^{- 1} (0.6) \approx 36.8 7^{\circ} \approx 36. 9^{\circ}$ .

$β = 9 0^{\circ} - 36.8 7^{\circ} = 53.1 3^{\circ} \approx 53. 1^{\circ}$ , because the angles of a triangle add up to 180°.

Check. From $β$ , the 6 cm side touches the angle, so it is the adjacent. So $cos β = 0.6$ , and $cos^{- 1} (0.6) \approx 53.1 3^{\circ}$ . ✓

Look at what the check showed. The same ratio, 0.6, is the sine of one angle and the cosine of the other. This is the link from the first trigonometry lecture: $cos θ = sin (9 0^{\circ} - θ)$ . The 6 cm side is opposite one angle and adjacent to the other, so the two ratios swap.

Worked Example 5 — Is this ladder safe?

A ladder is 5 m long. Its foot stands 2 m out from a wall. Find the angle between the ladder and the ground. Does it meet the "1 in 4" rule, which gives about 76°? If not, how far out should the foot be?

Think first. Draw the triangle. The ladder is the hypotenuse. The 2 m along the ground touches the angle at the foot, so it is the adjacent. Adjacent and hypotenuse means cosine.

cos θ = \frac{2}{5} = 0.4

θ = cos^{- 1} (0.4) \approx 66. 4^{\circ}

That is about 10° too shallow. A ladder at a shallow angle is more likely to slide out at the foot.

For the safe angle, use the rule itself: $tan θ = 4$ , so $θ = tan^{- 1} (4) \approx 75.9 6^{\circ}$ . The distance out is now an unknown side on top of the fraction, as in the last lecture: $d = 5 \times cos 75.9 6^{\circ} \approx 1.21$ m.

Check. The height up the wall is $5 \times sin 75.9 6^{\circ} \approx 4.85$ m, and $4.85 \div 1.21 \approx 4$ . So the foot is 1 out for every 4 up. ✓ The two lectures together give you both directions: sides from an angle, and an angle from sides.

Three traps

Trap 1 — Reading sin⁻¹ as "one over sine"

$sin^{- 1} (0.5) = 3 0^{\circ}$ , an angle. But $(sin 3 0^{\circ})^{- 1} = \frac{1}{sin 3 0 ^{\circ}} = 2$ , a number. The $- 1$ on the name of a function means "the inverse function". The $- 1$ on a number in brackets means "one over". Never use the calculator's $x^{- 1}$ key to find an angle.

Trap 2 — Rounding the ratio before the inverse

If you round $\frac{8}{13}$ to 0.62, then $sin^{- 1} (0.62) \approx 38. 3^{\circ}$ . The correct answer is 38.0°. A small change in the ratio becomes a noticeable change in the angle. Type the whole fraction into the calculator, inside the brackets.

Trap 3 — The wrong calculator mode

In radian mode, $sin^{- 1} (0.5)$ gives about 0.524, not 30. An angle of 0.524° in a triangle with sides 5 cm and 10 cm makes no sense, and that is the warning. Test the mode with $sin^{- 1} (0.5)$ before every exam.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A ramp rises 0.5 m over a horizontal run of 6 m. Find the angle between the ramp and the ground, to 1 d.p.

Hint 1
Which two sides do you know, from the angle at the bottom of the ramp? Is the ramp's own length given?

Hint 2
The rise is the opposite and the run is the adjacent, so $tan θ = \frac{0.5}{6}$ .

Full answer
$θ = tan^{- 1} (\frac{0.5}{6}) \approx 4.7 6^{\circ} \approx 4. 8^{\circ}$ (1 d.p.). Check: the rise is much shorter than the run, so the angle should be small, well below 45°. ✓ The ramp's length is $0. 5^{2} + 6^{2} \approx 6.021$ m, and $sin^{- 1} (0.5 \div 6.021) \approx 4.7 6^{\circ}$ agrees. ✓
A roof beam runs 4.5 m along the slope of a roof. Its top end is 1.8 m higher than its bottom end. Find the angle between the beam and the horizontal. A student gives 66.4°. What did the student do?

Hint 1
Draw the right-angled triangle. The beam is the hypotenuse. The 1.8 m rise is opposite the angle at the bottom.

Hint 2
$sin θ = \frac{1.8}{4.5} = 0.4$ . Now work out $cos^{- 1} (0.4)$ as well, and compare.

Full answer
$θ = sin^{- 1} (0.4) \approx 23.5 8^{\circ} \approx 23. 6^{\circ}$ . The student used $cos^{- 1} (0.4) \approx 66. 4^{\circ}$ . That is the angle at the top of the triangle, between the beam and the vertical, because from that angle the 1.8 m side is the adjacent. Check: $23.5 8^{\circ} + 66.4 2^{\circ} = 9 0^{\circ}$ . ✓ The rise is much shorter than the horizontal run, $4. 5^{2} - 1. 8^{2} \approx 4.12$ m, so the roof angle must be below 45°. ✓
A right-angled triangle has sides 7 cm, 24 cm and 25 cm. A student writes $sin θ = \frac{25}{7}$ for the smallest angle, and the calculator gives an error. Explain the error, then find both acute angles.

Hint 1
Which side is the hypotenuse? Can a sine be bigger than 1?

Hint 2
The smallest angle is opposite the shortest side, 7 cm. Sine is opposite ÷ hypotenuse.

Full answer
The student put the fraction upside down. $\frac{25}{7} \approx 3.57$ , and no sine can be bigger than 1, so there is no angle. The hypotenuse, 25 cm, goes on the bottom: $sin θ = \frac{7}{25} = 0.28$ , so $θ = sin^{- 1} (0.28) \approx 16.2 6^{\circ} \approx 16. 3^{\circ}$ . The other angle is $9 0^{\circ} - 16.2 6^{\circ} \approx 73. 7^{\circ}$ . Check: $tan^{- 1} (\frac{24}{7}) \approx 73.7 4^{\circ}$ . ✓ The smallest angle is opposite the smallest side, as it should be. ✓
A stretch. A rectangle is 8 cm long and 5 cm wide. Find the angle between a diagonal and a long side. Then find both angles between the two diagonals, where they cross.

Hint 1
A diagonal cuts the rectangle into two right-angled triangles with shorter sides 8 cm and 5 cm.

Hint 2
The diagonals cut each other in half, so they form isosceles triangles. In the triangle with the long side as its base, both base angles equal your first answer.

Full answer
From the corner, the 5 cm side is opposite and the 8 cm side is adjacent: $θ = tan^{- 1} (\frac{5}{8}) \approx 32.0 1^{\circ} \approx 32. 0^{\circ}$ . The diagonals have equal length and cut each other in half, so the triangle on a long side has two equal sides. Its base angles are both 32.01°. The angle at the crossing point is $18 0^{\circ} - 2 \times 32.0 1^{\circ} \approx 115.9 9^{\circ} \approx 116. 0^{\circ}$ . The angles on a straight line add up to 180°, so the other angle between the diagonals is $18 0^{\circ} - 115.9 9^{\circ} \approx 64. 0^{\circ}$ . Check: 64.0° is twice 32.0°, which fits, because it is the apex angle of the triangle on a short side, whose base angles are $9 0^{\circ} - 32.0 1^{\circ} \approx 57.9 9^{\circ}$ , and $18 0^{\circ} - 2 \times 57.9 9^{\circ} \approx 64. 0^{\circ}$ . ✓

The bigger idea

Long before calculators, people found angles from ratios by reading printed tables of sines backwards. They looked down the column of sines for the ratio they had, then read the angle beside it, and estimated between two rows when the ratio fell between them. That is exactly the search you did by hand at the start of this lecture. The inverse button is the same idea, made fast.

There is one more thing to notice. In a right-angled triangle, every angle is between 0° and 90°, and each ratio belongs to exactly one of them. Beyond 90°, this stops being true. The inverse functions lecture said that many angles have the same sine, so the calculator always chooses its answer from one restricted range. For example, $sin 15 0^{\circ}$ is also 0.5. The sine rule meets this problem in triangles with an obtuse angle, and the unit circle at Pre-University explains it fully. Inside a right-angled triangle, you can trust the calculator's single answer.

Key concepts in this lesson

When two sides of a right-angled triangle are known and an angle is wanted, the unknown is inside the ratio, so you need the inverse function.
$sin^{- 1} (x)$ is the angle whose sine is $x$ , and the same is true for $cos^{- 1}$ and $tan^{- 1}$ .
Choose the inverse from the pair of known sides, exactly as you choose the ratio, and type the whole fraction into the calculator.
The $- 1$ in $sin^{- 1}$ names the inverse function and does not mean "one over sine".
The sine and cosine of an angle in a right-angled triangle are less than 1, so $sin^{- 1}$ or $cos^{- 1}$ of a number bigger than 1 has no answer.