Geometry · G.SRT.C.8

Inverse trigonometric ratios

You know two sides of a right triangle but neither of its acute angles. How do you get the angle back from the sides?

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Sine, cosine and tangentGeometry

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Magnitude and direction of a vectorPrecalculus Law of cosinesPrecalculus

Where you are 7 in Right triangle trigonometry

Trigonometry7

Right triangle trigonometry7

Opposite, adjacent and hypotenuse Sine, cosine and tangent

Inverse trigonometric ratios

Secant, cosecant and cotangent ratios Solving right triangles ▾ Show all 7 (2 more)

When you reach for it

Solving right triangles starts from an angle and a side and finds the other sides. Very often the situation is the other way round. A ramp rises 0.5 meters over 6 meters of floor; a rectangle is 9 wide and 4 high; a boat has gone 3 km east and 7 km north. Lengths are easy to measure, and two of them are known. The angle is what is wanted.

Two sides of a right triangle fix it completely: two right triangles with the same two legs, or the same leg and hypotenuse, are congruent. So the acute angles are fixed too, and the only question is how to compute them. The two known sides give a ratio, and the ratio belongs to the angle, as sine, cosine and tangent showed. This lesson runs that link backwards: from the ratio to the angle.

The procedure

In the figure the right angle is at C, the leg AC is 8, the leg BC is 5, and the angle \theta at A is wanted.

Figure 1Two legs are known and the angle \theta at A is wanted.

Name the two known sides from the wanted angle

The names opposite, adjacent and hypotenuse depend on which angle you stand at, as Opposite, adjacent and hypotenuse set out, so stand at the angle you want. From A, the leg BC = 5 is opposite and the leg AC = 8 is adjacent. The hypotenuse AB is not known.

Choose the ratio that holds both

Each ratio uses exactly two of the three sides: sine uses opposite and hypotenuse, cosine uses adjacent and hypotenuse, tangent uses opposite and adjacent. Pick the one whose two sides are the two you know, so that the equation has no unknown in it except the angle. Here that is the tangent:

\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{8} = 0.625

Undo the ratio

Now the question is: which acute angle has tangent 0.625? There is exactly one. As an acute angle opens from 0^\circ toward 90^\circ, its sine and its tangent grow and its cosine shrinks, all without ever turning back, so no value is taken twice. One value of the ratio means one angle.

That angle gets a name. For a number x between 0 and 1, \sin^{-1} x is the acute angle whose sine is x, and \cos^{-1} x is the acute angle whose cosine is x. For any positive number x, \tan^{-1} x is the acute angle whose tangent is x. They are read “inverse sine”, “inverse cosine” and “inverse tangent”, and some calculators label them \text{arcsin}, \text{arccos} and \text{arctan}. In symbols,

\sin \theta = x \quad\Longleftrightarrow\quad \theta = \sin^{-1} x,

for an acute angle \theta, and the same with cosine and tangent. The -1 is not a power: \sin^{-1} x is an angle, while \displaystyle (\sin x)^{-1} = \frac{1}{\sin x} is a ratio of lengths, which has its own name in secant, cosecant and cotangent.

So in the triangle,

\theta = \tan^{-1}\!\left(\frac{5}{8}\right) \approx 32.0^\circ.

A calculator in degree mode gives 32.005\ldots, rounded here to the nearest tenth of a degree. As a check, \tan 32.0^\circ \approx 0.625, which is the ratio we started from.

The other acute angle

The three angles of a triangle add to 180^\circ and the right angle takes 90^\circ, so the two acute angles add to 90^\circ:

\angle B = 90^\circ - \theta \approx 58.0^\circ.

It can also be found on its own. From B the leg AC = 8 is opposite and BC = 5 is adjacent, so \displaystyle \angle B = \tan^{-1}\left(\frac{8}{5}\right) \approx 58.0^\circ, the same answer by a separate route, which is a good check.

Use the sides you were given

If the third side is wanted as well, the Pythagorean theorem gives it: \displaystyle AB = \sqrt{8^2 + 5^2} = \sqrt{89} \approx 9.43. Any two of the three sides now give \theta, but use the two that were given. A side you computed is already rounded, and an angle found from it carries that rounding with it.

Worked examples

Angles are given to the nearest tenth of a degree and lengths to two decimal places, unless the example says otherwise. Every calculator here is in degree mode.

Example 1

Reading the three inverses

Find each angle and say what it means: (a) \sin^{-1}(0.6); (b) \cos^{-1}(0.25); (c) \tan^{-1}(3).

Solution

Each one asks for an acute angle, and the calculator’s inverse key finds it.

(a) \sin^{-1}(0.6) \approx 36.9^\circ. This is the acute angle whose opposite side is 0.6 of the hypotenuse. Checking it the other way, \sin 36.9^\circ \approx 0.6004; the small difference is the rounding of the angle. A calculator left in radian mode would show 0.6435 instead, which is the same angle in a different unit, so check the mode first.

(b) \cos^{-1}(0.25) \approx 75.5^\circ. An adjacent side only a quarter of the hypotenuse means the angle is large, close to 90^\circ, and it is.

(c) \tan^{-1}(3) \approx 71.6^\circ. A tangent of 1 means equal legs and a 45^\circ angle; a tangent of 3 means the opposite leg is three times the adjacent one, so the angle is well past 45^\circ.

Example 2

Choosing the ratio from the two sides

Each part gives two sides of a right triangle, named from an acute angle \theta. Find \theta.

(a) The side opposite \theta is 7 and the hypotenuse is 11.

(b) The side adjacent to \theta is 10 and the hypotenuse is 13.

Solution

The two known sides decide the ratio, and the ratio decides the inverse.

(a) Opposite and hypotenuse are the sides of the sine:

\begin{aligned} \sin \theta &= \frac{7}{11} \\ \theta &= \sin^{-1}\!\left(\frac{7}{11}\right) \approx 39.5^\circ \end{aligned}

(b) Adjacent and hypotenuse are the sides of the cosine:

\begin{aligned} \cos \theta &= \frac{10}{13} \\ \theta &= \cos^{-1}\!\left(\frac{10}{13}\right) \approx 39.7^\circ \end{aligned}

\begin{aligned} \tan \theta &= \frac{5}{3} \\ \theta &= \tan^{-1}\!\left(\frac{5}{3}\right) \approx 59.0^\circ \end{aligned}

In (c) the opposite leg is the longer leg, so \theta is the larger acute angle and must be more than 45^\circ, which it is. Typing the fraction straight into the calculator, rather than a rounded decimal for it, keeps every digit.

Example 3

Solving the whole triangle from two sides

In the right triangle PQR the right angle is at R, the hypotenuse PQ is 16 and the leg QR is 6. Find PR, \angle P and \angle Q.

Figure 2Example 3. A leg and the hypotenuse are known; both acute angles and the third side are wanted.

Solution

The third side comes first from the Pythagorean theorem, since it needs no angle:

PR = \sqrt{16^2 - 6^2} = \sqrt{220} \approx 14.83

For \angle P, the side QR = 6 is opposite and PQ = 16 is the hypotenuse, so the sine holds both:

\begin{aligned} \sin P &= \frac{6}{16} = 0.375 \\ \angle P &= \sin^{-1}(0.375) \approx 22.0^\circ \end{aligned}

For \angle Q, the same side QR = 6 is now adjacent, so the cosine holds both given sides:

\begin{aligned} \cos Q &= \frac{6}{16} \\ \angle Q &= \cos^{-1}(0.375) \approx 68.0^\circ \end{aligned}

The same ratio gives both angles, through the sine at one and the cosine at the other, because a side opposite one acute angle is adjacent to the other. The two answers add to 90^\circ, as they must. A last check uses the computed side: \displaystyle \tan^{-1}\left(\frac{6}{\sqrt{220}}\right) is again 22.0^\circ.

Example 4

Which side is 6 meters?

A ramp rises 0.5 m. Find the angle it makes with the floor, to two decimal places of a degree, (a) if it covers 6 m of floor, and (b) if the sloping surface of the ramp is 6 m long.

Solution

The rise, the floor it covers and the sloping surface form a right triangle with the right angle under the top of the ramp. The angle \theta is at the foot of the ramp, so the rise is opposite it.

(a) The 6 m of floor is the leg adjacent to \theta. Opposite and adjacent call for the tangent:

\theta = \tan^{-1}\!\left(\frac{0.5}{6}\right) \approx 4.76^\circ

(b) Now 6 m is the sloping surface, the hypotenuse. Opposite and hypotenuse call for the sine:

\theta = \sin^{-1}\!\left(\frac{0.5}{6}\right) \approx 4.78^\circ

The two answers are different, because the triangles are: in (b) the floor covered is only \displaystyle \sqrt{6^2 - 0.5^2} \approx 5.98 m, so the same rise happens over less floor and the ramp is a little steeper. For a shallow ramp the difference is small, but it is never zero, and “a side of 6” means nothing until it says which side.

Example 5

When the ratio is one you know

Find \theta exactly. (a) The hypotenuse is 14 and the side opposite \theta is 7. (b) The two legs are both 6.

Solution

(a) \displaystyle \sin \theta = \frac{7}{14} = \frac{1}{2}. The sine of 30^\circ is exactly \displaystyle \frac{1}{2}, so \displaystyle \theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ with no rounding at all.

(b) \displaystyle \tan \theta = \frac{6}{6} = 1. Equal legs make an isosceles right triangle, whose acute angles are equal and add to 90^\circ, so \theta = \tan^{-1}(1) = 45^\circ exactly.

When the ratio is one of the values worked out in trig values of the special right triangles, the angle is exact; otherwise it comes from a calculator and is rounded.

Example 6

The angle between the diagonals of a rectangle

A rectangle ABCD is 9 wide and 4 high. Find the angle \theta that the diagonal AC makes with the long side AB, and the acute angle \alpha at which the two diagonals cross.

Figure 3Example 6. The diagonal makes the angle \theta with the long side; the diagonals cross at the angle \alpha.

Solution

The diagonal AC cuts the rectangle into two right triangles, and in triangle ABC the right angle is at B. From A, the side BC = 4 is opposite and AB = 9 is adjacent, so

\theta = \tan^{-1}\!\left(\frac{4}{9}\right) \approx 24.0^\circ.

The diagonals of a rectangle are equal and cut each other in half, so MB = MC, and triangle MBC is isosceles with base BC. Its base angle \angle MBC is \angle DBC, an acute angle of the right triangle BCD, whose right angle is at C. From B, the side DC = 9 is opposite and BC = 4 is adjacent, so \displaystyle \angle MBC = \tan^{-1}\left(\frac{9}{4}\right). That is the other acute angle of a right triangle with legs 9 and 4, so it is 90^\circ - \theta. The base angles of an isosceles triangle are equal, so

\begin{aligned} \alpha &= 180^\circ - 2(90^\circ - \theta) \\ &= 2\theta \approx 47.9^\circ. \end{aligned}

Doubling the unrounded \theta = 23.96\ldots^\circ gives 47.9^\circ; doubling the rounded 24.0^\circ would give 48.0^\circ, one more reason to keep every digit until the end. The diagonals also make the angle 180^\circ - \alpha \approx 132.1^\circ, the one facing the long sides. Triangle MBC has no right angle of its own; the inverse tangent worked because a right triangle sharing one of its angles was found first.

Example 7

A direction from two movements

A boat sails 3 km due east and then 7 km due north. How far is it from where it started, and what angle does the straight line back to the start make with due north?

Solution

East and north are at right angles, so the two legs of the trip and the straight line joining its ends form a right triangle with legs 3 and 7. The straight-line distance is the hypotenuse:

d = \sqrt{3^2 + 7^2} = \sqrt{58} \approx 7.62 \text{ km}

The angle wanted is at the start, between the hypotenuse and the north direction. Measured from north, the 7 km north leg is adjacent and the 3 km east leg is opposite, so

\theta = \tan^{-1}\!\left(\frac{3}{7}\right) \approx 23.2^\circ.

The line runs 23.2^\circ east of due north. The angle it makes with due east is the other acute angle, \displaystyle \tan^{-1}\left(\frac{7}{3}\right) \approx 66.8^\circ, and the two add to 90^\circ. A direction like this, an angle from north, is the start of bearings.

Boundaries

There must be a right angle. The ratios, and so their inverses, are defined from a right triangle. Two sides of a triangle with no right angle do not fix it at all, since the angle between them can be anything; three sides do, and the law of cosines then finds its angles. Example 6 shows the other way round a missing right angle: find a right triangle that shares the angle.

The sides must be named from the angle. “A side of 6” could be either leg or the hypotenuse, and Example 4 shows that each gives a different angle.

A sine or cosine ratio must lie strictly between 0 and 1. A leg is always shorter than the hypotenuse, so in a right triangle \displaystyle \frac{\text{opposite}}{\text{hypotenuse}} is less than 1. If a calculation asks for \sin^{-1}(1.3), the calculator will report an error, and the error is a message: the side called the hypotenuse is not the longest side, so the sides were named wrongly or cannot belong to a right triangle. A tangent ratio can be any positive number, since either leg can be the longer.

These inverses return only acute angles, here. Calculators also accept 0, negative numbers and 1, and return angles such as 0^\circ, 90^\circ or negative angles. Those belong to sine and cosine defined for angles of every size, and why each inverse then returns only one chosen slice of angles is the subject of inverse trig functions.

The answer is rounded, except for a few special ratios. For most ratios the angle is irrational in degrees and comes from a calculator. Exact angles appear for special values such as \displaystyle \frac{1}{2}, \displaystyle \frac{\sqrt{2}}{2}, \displaystyle \frac{\sqrt{3}}{2} and 1, as in Example 5. Round once, at the end.

Degrees and radians are different units. The inverse keys answer in whichever unit the calculator is set to: \sin^{-1}(0.6) is 36.87^\circ in degree mode and 0.6435 in radian mode.

Where this goes next

Together with solving right triangles this finishes the right triangle: any two sides, or a side and an acute angle, now give everything else. When a height or distance cannot be reached from one place, sightings from two places pin it down, which is the next lesson.

Example 7 is the direction of a movement given by its east and north parts. Written as a pair (3, 7), that movement is a vector, and its length and its angle, found exactly as here, are its magnitude and direction.

For a triangle with no right angle, three sides fix the angles, and the law of cosines turns the three sides into the cosine of each angle; an inverse cosine then finishes the job, once it is widened to accept negative values for obtuse angles.

Finally, once sine and cosine are defined for every angle on the unit circle, a value such as \displaystyle \frac{1}{2} is the sine of infinitely many angles, not one. Choosing which one \sin^{-1} should return is the business of inverse trig functions, which grow out of the inverse functions of algebra.

Exercises

Give angles to the nearest tenth of a degree and lengths to two decimal places, unless the exercise asks for an exact answer.

Evaluate the inverse

\sin^{-1}(0.35)
\cos^{-1}(0.71)
\tan^{-1}(0.3)
\tan^{-1}(12)
\displaystyle \sin^{-1}\left(\frac{2}{9}\right)

Find the angle from two sides

In each, \theta is an acute angle of a right triangle. Find \theta.

The side opposite \theta is 5 and the hypotenuse is 9.
The side adjacent to \theta is 11 and the hypotenuse is 14.
The side opposite \theta is 13 and the side adjacent to it is 6.
The side opposite \theta is 1.8 and the side adjacent to it is 2.5.
The side adjacent to \theta is 3 and the hypotenuse is 20.

Exact angles

Find \theta exactly, without a calculator.

The hypotenuse is 10 and the side opposite \theta is 5.
The hypotenuse is \displaystyle 6\sqrt{2} and the side opposite \theta is 6.
The hypotenuse is 8 and the side adjacent to \theta is 4.

Solve the triangle

In each, the right angle of triangle ABC is at C. Find the third side and both acute angles.

AC = 20 and BC = 21.
AB = 25 and AC = 7.
AB = 13.5 and BC = 4.2.

Problems

A kite flies at the end of a straight string 40 m long, 28 m above the hand holding the string. Find the angle the string makes with the horizontal.
A road climbs 45 m over a horizontal distance of 800 m. Find the angle the road makes with the horizontal.
A rectangle is 15 by 8. Find the angle a diagonal makes with a longer side, and the acute angle between the two diagonals.
An isosceles triangle has two sides of 9 and a base of 14. Find all three of its angles.
The legs of a right triangle are in the ratio 2 : 5. Find its acute angles.

Reasoning

Explain why no acute angle has sine 1.3, and what it tells you when a calculation of an angle in a right triangle arrives at \sin^{-1}(1.3).
Show that for every number x with 0 < x < 1, \sin^{-1} x + \cos^{-1} x = 90^\circ.
Ilse reads \sin^{-1}(0.5) as \displaystyle \frac{1}{\sin 0.5}. Explain what each of the two expressions means, and why they cannot be the same thing.
One right triangle has legs 3 and 5, another has legs 6 and 10. Explain, without finding the angles, why the two triangles have the same acute angles.

Answers

20.5^\circ. The acute angle whose sine is 0.35.
44.8^\circ. Just under 45^\circ, since \cos 45^\circ \approx 0.707 is a little less than 0.71 and the cosine shrinks as the angle grows.
16.7^\circ.
85.2^\circ. A tangent of 12 means the opposite leg is twelve times the adjacent one, so the angle is close to 90^\circ.
12.8^\circ.
33.7^\circ. Opposite and hypotenuse: \displaystyle \theta = \sin^{-1}\left(\frac{5}{9}\right).
38.2^\circ. Adjacent and hypotenuse: \displaystyle \theta = \cos^{-1}\left(\frac{11}{14}\right).
65.2^\circ. Opposite and adjacent: \displaystyle \theta = \tan^{-1}\left(\frac{13}{6}\right).
35.8^\circ. Opposite and adjacent: \displaystyle \theta = \tan^{-1}\left(\frac{1.8}{2.5}\right).
81.4^\circ. Adjacent and hypotenuse: \theta = \cos^{-1}(0.15).
30^\circ. \displaystyle \sin \theta = \frac{5}{10} = \frac{1}{2}, and \displaystyle \sin 30^\circ = \frac{1}{2}.
45^\circ. \displaystyle \sin \theta = \frac{6}{6\sqrt{2}} = \frac{\sqrt{2}}{2}, the sine of 45^\circ. The other leg is \displaystyle \sqrt{72 - 36} = 6, so the legs are equal, as they must be.
60^\circ. \displaystyle \cos \theta = \frac{4}{8} = \frac{1}{2}, and \displaystyle \cos 60^\circ = \frac{1}{2}.
AB = 29, \angle A \approx 46.4^\circ, \angle B \approx 43.6^\circ. \displaystyle AB = \sqrt{20^2 + 21^2} = \sqrt{841} = 29. From A, BC is opposite and AC adjacent, so \displaystyle \angle A = \tan^{-1}\left(\frac{21}{20}\right); \displaystyle \angle B = \tan^{-1}\left(\frac{20}{21}\right).
BC = 24, \angle A \approx 73.7^\circ, \angle B \approx 16.3^\circ. \displaystyle BC = \sqrt{25^2 - 7^2} = \sqrt{576} = 24. AC is adjacent to A, so \displaystyle \angle A = \cos^{-1}\left(\frac{7}{25}\right); it is opposite B, so \displaystyle \angle B = \sin^{-1}\left(\frac{7}{25}\right).
AC \approx 12.83, \angle A \approx 18.1^\circ, \angle B \approx 71.9^\circ. \displaystyle AC = \sqrt{13.5^2 - 4.2^2} = \sqrt{164.61}. BC is opposite A, so \displaystyle \angle A = \sin^{-1}\left(\frac{4.2}{13.5}\right); it is adjacent to B, so \displaystyle \angle B = \cos^{-1}\left(\frac{4.2}{13.5}\right).
44.4^\circ. The string is the hypotenuse and the height is opposite the angle at the hand: \displaystyle \sin^{-1}\left(\frac{28}{40}\right) = \sin^{-1}(0.7).
3.2^\circ. The climb is opposite and the horizontal distance adjacent: \displaystyle \tan^{-1}\left(\frac{45}{800}\right).
28.1^\circ and 56.1^\circ. The diagonal and two sides make a right triangle, and from the corner the short side 8 is opposite and the long side 15 adjacent, so the angle is \displaystyle \tan^{-1}\left(\frac{8}{15}\right) \approx 28.07^\circ. As in Example 6, the diagonals cross at twice this angle, 56.1^\circ, doubled before rounding.
Base angles 38.9^\circ each, top angle 102.1^\circ. The height from the top meets the base at its midpoint and makes two right triangles with hypotenuse 9 and a leg of 7 along the base. That leg is adjacent to a base angle, so each base angle is \displaystyle \cos^{-1}\left(\frac{7}{9}\right) \approx 38.94^\circ, and the top angle is 180^\circ - 2 \times 38.94\ldots^\circ, from the unrounded value.
21.8^\circ and 68.2^\circ. Call the legs 2k and 5k. The angle opposite the shorter leg is \displaystyle \tan^{-1}\left(\frac{2k}{5k}\right) = \tan^{-1}(0.4), whatever k is; the other is \tan^{-1}(2.5).
For an acute angle in a right triangle, the sine is a leg divided by the hypotenuse, and the hypotenuse is the longest side, so the sine is less than 1. No acute angle has sine 1.3. A calculation reaching \sin^{-1}(1.3) has divided a longer side by a shorter one: the side used as the hypotenuse is not the longest, so either the sides were named from the wrong angle or the three lengths cannot form a right triangle.
Draw a right triangle with hypotenuse 1 and one leg x, which is possible because 0 < x < 1. Call the acute angle opposite that leg \alpha and the other one \beta. Then \displaystyle \sin \alpha = \frac{x}{1} = x, so \alpha = \sin^{-1} x. The same leg is adjacent to \beta, so \cos \beta = x and \beta = \cos^{-1} x. The two acute angles of a right triangle add to 90^\circ, so \sin^{-1} x + \cos^{-1} x = \alpha + \beta = 90^\circ.
\sin^{-1}(0.5) is an angle: the acute angle whose sine is 0.5, which is 30^\circ. The expression \displaystyle \frac{1}{\sin 0.5} starts from an angle of size 0.5, takes its sine, which is a ratio of two lengths, and turns that ratio upside down, which gives another ratio of lengths (the cosecant of the angle 0.5). One is an angle and the other is a pure number, so they are different kinds of thing. The -1 in \sin^{-1} means “undo the sine”, not “one over”.
In both triangles the tangent of the angle opposite the shorter leg is the same number, \displaystyle \frac{3}{5} = \frac{6}{10}, and only one acute angle has that tangent. So that angle is the same in both, and the other acute angle, 90^\circ minus it, is the same too. Put another way, the second triangle is the first scaled by 2, and scaling a figure does not change its angles (similarity).