Trigonometry of the right triangle

Nothing built so far repeats. Polynomials, exponentials and logarithms all eventually head off in one direction and stay there, and yet a tide, a note and a rotating shaft all come back to where they started, over and over.

The functions that repeat come from angles, and this lesson builds them where they were first built, inside a right triangle. It assumes the algebra of the earlier lessons and Pythagoras' theorem, and it ends with two rules that solve any triangle at all, including those with no right angle in them.

One angle fixes every ratio

Take a right triangle with one acute angle $θ$ . Now draw a bigger one with the same angle $θ$ . The two triangles have the same three angles, since the right angles match and the third is whatever is left of $180 °$ , so they are similar: one is a scaled copy of the other.

Similar triangles have proportional sides. Doubling every side of the first triangle gives the second, or tripling, or multiplying by $1.7$ , and in every case the ratio of any two sides within a triangle is unchanged, because both members of the ratio were multiplied by the same factor. This is the fact everything below rests on: the ratios depend on the angle alone, not on the size of the triangle.

So the ratios can be named as functions of the angle. Label the side opposite $θ$ as opposite, the one next to it that is not the hypotenuse as adjacent, and define

\sin θ = \frac{opposite}{hypotenuse}, \cos θ = \frac{adjacent}{hypotenuse}, \tan θ = \frac{opposite}{adjacent}

Dividing the first by the second cancels the hypotenuse and leaves the third, so $\tan θ = \sin θ / \cos θ$ always, which is one fewer thing to remember. Because the hypotenuse is the longest side, sine and cosine of an acute angle always lie strictly between $0$ and $1$ , while the tangent is unbounded and grows without limit as $θ$ approaches $90 °$ .

Two triangles worth knowing exactly

Most values of these functions are irrational and are looked up or computed. Two triangles give exact values, and they are worth knowing because they appear constantly.

Cut a square of side $1$ along its diagonal. The result is a right triangle with two legs of $1$ and, by Pythagoras, a hypotenuse of $\sqrt{2}$ , and its acute angles are both $45 °$ . So $\sin 45 ° = \cos 45 ° = 1 / \sqrt{2} = 0.707107$ and $\tan 45 ° = 1$ .

Cut an equilateral triangle of side $2$ down the middle. The result is a right triangle with hypotenuse $2$ , short leg $1$ , and remaining leg $\sqrt{4 - 1} = \sqrt{3}$ , with angles of $30 °$ and $60 °$ . Reading the ratios off gives $\sin 30 ° = 1 / 2$ , $\cos 30 ° = \sqrt{3} / 2 = 0.866025$ , $\tan 30 ° = 1 / \sqrt{3} = 0.577350$ , and the same three with sine and cosine swapped at $60 °$ . The swap is not a coincidence: the two acute angles of a right triangle add to $90 °$ , and one angle's opposite side is the other's adjacent, so $\sin θ = \cos (90 ° - θ)$ for every acute $θ$ . That relation is where the word cosine comes from, the sine of the complement.

Pythagoras' theorem itself becomes an identity in this language. In a right triangle with hypotenuse $1$ , the legs are exactly $\cos θ$ and $\sin θ$ , so $a^{2} + b^{2} = c^{2}$ reads

\sin^{2} θ + \cos^{2} θ = 1

Checking at $37 °$ : ${0.60182}^{2} + {0.79864}^{2} = 0.36219 + 0.63781 = 1$ . This is the most used identity in the subject, and it is Pythagoras wearing different clothes.

Solving a right triangle

Given one side and one acute angle, the other two sides follow. The discipline is to write the ratio that connects what you know to what you want, then rearrange.

Example. Standing $24$ m from the base of a tree, an observer whose eye is $1.6$ m above the ground measures the angle of elevation of the top as $37 °$ . How tall is the tree?

The horizontal distance is adjacent to the angle and the height above eye level is opposite, so the tangent connects them: $\tan 37 ° = h / 24$ , giving $h = 24 \times 0.75355 = 18.09$ m above eye level. The tree is $18.09 + 1.6 = 19.69$ m tall. The eye height matters: omitting it understates the tree by more than eight per cent.

Now you. A ladder $5.2$ m long leans against a wall at $68 °$ to the horizontal. How far up the wall does it reach, and how far is its foot from the wall?

Answer

The reach is opposite the angle, so it is $5.2 \sin 68 ° = 5.2 \times 0.92718 = 4.821$ m, and the foot is $5.2 \cos 68 ° = 5.2 \times 0.37461 = 1.948$ m from the wall.

Going the other way, from a ratio to the angle, needs the inverse functions $\sin^{- 1}$ , $\cos^{- 1}$ and $\tan^{- 1}$ , also written arcsin, arccos and arctan. Here the restriction from the fourth lesson bites. Sine is not injective over all angles, so it has no inverse until its domain is cut down, and the convention takes $\sin^{- 1}$ to return an angle between $- 90 °$ and $90 °$ . That is a choice, and it is the reason a calculator answers $\sin^{- 1} (0.5)$ with $30 °$ and never mentions $150 °$ , which has the same sine. Anyone solving a real problem must decide whether the other angle is the one wanted.

A road signed as a gradient of $1$ in $8$ rises one metre for every eight along, so its angle is $\tan^{- 1} (0.125) = 7.13 °$ . Signs quoted as percentages mean the same thing: a $12$ per cent grade is $\tan^{- 1} (0.12) = 6.84 °$ . Road gradients are always small enough that the angle in radians, the tangent and the sine agree to within a fraction of a per cent, which is why the sloppiness rarely matters in practice and matters completely on a roof or a ramp.

Reaching triangles with no right angle

Real triangles are rarely right angled, and the definitions above do not apply to them directly. They can be made to apply by dropping a perpendicular, and doing so produces two rules that cover every case.

Label a triangle with angles $A$ , $B$ , $C$ and the side opposite each with the matching lower-case letter. Drop the altitude $h$ from $C$ to the side $c$ . It splits the triangle into two right triangles, and in each of them $h$ is opposite a known angle: $h = b \sin A$ from one, $h = a \sin B$ from the other. Setting these equal gives $b \sin A = a \sin B$ , that is

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

the third ratio following by dropping a different altitude. This is the sine rule, and it solves a triangle whenever an angle and its opposite side are both known.

Example. A triangle has $A = 41 °$ , $B = 63 °$ and $a = 12$ . Find the remaining angle and sides.

The angles sum to $180 °$ , so $C = 76 °$ . Then $b = a \sin B / \sin A = 12 \times 0.89101 / 0.65606 = 16.297$ and $c = a \sin C / \sin A = 12 \times 0.97030 / 0.65606 = 17.748$ . A check on plausibility: the largest side should face the largest angle, and it does.

Now you. A triangle has $A = 38 °$ , $B = 57 °$ and $a = 9$ . Find $C$ , $b$ and $c$ .

Answer

$C = 85 °$ . Then $b = 9 \times 0.83867 / 0.61566 = 12.26$ and $c = 9 \times 0.99619 / 0.61566 = 14.563$ .

The sine rule has a trap that must be stated. Given two sides and an angle not between them, the rule can produce two valid triangles. With $a = 7$ , $b = 10$ and $A = 40 °$ , it gives $\sin B = 10 \sin 40 ° / 7 = 0.9183$ , and both $B = 66.67 °$ and $B = 113.33 °$ satisfy that, since supplementary angles have equal sines. Each leads to a genuine triangle, one with $C = 73.31 °$ and one with $C = 26.69 °$ . This is the ambiguous case, and no amount of algebra resolves it: the data really do describe two shapes, and only extra information settles which.

The cosine rule

When no angle and its opposite side are both known, the sine rule cannot start. Two cases fall outside it: three sides given, or two sides and the angle between them. The cosine rule covers both:

c^{2} = a^{2} + b^{2} - 2 a b \cos C

Its derivation is the same altitude trick. Drop the perpendicular from $B$ to the side $b$ , splitting it into pieces of length $a \cos C$ and $b - a \cos C$ , with height $a \sin C$ . Pythagoras on the right-hand triangle gives $c^{2} = (a \sin C)^{2} + (b - a \cos C)^{2}$ , and expanding gives $a^{2} \sin^{2} C + b^{2} - 2 a b \cos C + a^{2} \cos^{2} C$ . The two squared trigonometric terms combine to $a^{2}$ by the Pythagorean identity, leaving the rule.

Set $C = 90 °$ and the cosine term vanishes, recovering $c^{2} = a^{2} + b^{2}$ . So Pythagoras is the special case, and the $- 2 a b \cos C$ is the correction for the angle being something other than a right angle.

Example. Two sides of a triangle are $7$ and $9$ with an included angle of $52 °$ . Find the third side and the area.

$c^{2} = 49 + 81 - 2 \times 63 \times \cos 52 ° = 130 - 126 \times 0.61566 = 130 - 77.57 = 52.43$ , so $c = 7.241$ . The area of any triangle is half the product of two sides and the sine of the angle between them, which follows from taking one side as the base and $a \sin C$ as the height: area $= 0.5 \times 63 \times \sin 52 ° = 0.5 \times 63 \times 0.78801 = 24.82$ .

Now you. Two sides are $11$ and $6$ with an included angle of $78 °$ . Find the third side and the area.

Answer

$c^{2} = 121 + 36 - 132 \cos 78 ° = 157 - 132 \times 0.20791 = 129.56$ , so $c = 11.382$ . The area is $0.5 \times 66 \times \sin 78 ° = 0.5 \times 66 \times 0.97815 = 32.28$ .

Rearranged, the cosine rule finds an angle from three sides: $\cos C = (a^{2} + b^{2} - c^{2}) / 2 a b$ . For sides $5$ , $7$ and $10$ the angle opposite the longest is $\cos^{- 1} big ((25 + 49 - 100) / 70 big) = \cos^{- 1} (- 0.3714) = 111.8 °$ , and the negative cosine correctly signals an obtuse angle. Unlike the sine rule, this version is never ambiguous, because cosine takes each value once between $0 °$ and $180 °$ .

Where these definitions stop

Everything in this lesson has been about an angle inside a right triangle, which means an angle strictly between $0 °$ and $90 °$ . Ask for $\sin 150 °$ and the definitions give nothing: no right triangle has an angle of $150 °$ in it, so there is no opposite side to divide by a hypotenuse.

Yet the ambiguous case just used $\sin 113.33 °$ , and the cosine rule just used $\cos 111.8 °$ , both of which were treated as ordinary numbers. Something has been assumed that has not been defined, and that is a debt to settle rather than a detail to skip.

The repair is to leave the triangle behind and put the angle at the centre of a circle, where it can be as large as you like and can turn as many times as it likes. That construction gives sine and cosine a value for every real number, produces the repeating functions this course still owes, and is the next lesson.