Sine, cosine and tangent: from angle to height

You know how far you stand from a post and the angle at which you see its top. Pythagoras’ theorem is not enough yet: you have only one length. Trigonometry supplies the connection to the angle.

For an acute angle in a right triangle, sine is opposite leg divided by hypotenuse; cosine is adjacent leg divided by hypotenuse; tangent is opposite leg divided by adjacent leg. These ratios depend on the angle, rather than triangle size. The reason is AA similarity; Pythagoras’ theorem helps check the lengths.

Choose an angle, name the sides

Let α be the chosen acute angle, a its opposite leg, b its adjacent leg, and c the hypotenuse. For 0°<α<90°:

$\sin\alpha=\frac{a}{c},\qquad \cos\alpha=\frac{b}{c},\qquad \tan\alpha=\frac{a}{b}.$

The hypotenuse is always opposite the right angle. It touches vertex α but is not the adjacent leg. Choosing the other acute angle swaps the roles of a and b; the hypotenuse stays the same. In the diagram sinα=3/5, cosα=4/5 and tanα=3/4. AMSI on definitions and side choice.

Sides relative to α

Measure the angle from horizontal

0.6 / 3 = 20%, not 20°

Each panel has its own uniform scale. First: α is at A; opposite BC=a=3, adjacent AB=b=4, hypotenuse AC=c=5. Second: horizontal d=12 m is drawn at instrument height e=1.6 m; a 35° sight line gives H≈10.0 m when both ground levels agree. Third: h=0.6 m,d=3 m,L≈3.06 m, angle≈11.31°; no vertical exaggeration. Right-angle marks distinguish height from slope.

All three ratios are dimensionless: metres divide by metres. For an acute angle, 0<sinα<1, 0<cosα<1 and tanα>0. Doubling every side changes none of the ratios.

Let the data choose the formula

You need not recall every formula at once: mark the known side and the unknown one.

Opposite leg and hypotenuse → sine: a=c sinα or c=a/sinα.
Adjacent leg and hypotenuse → cosine: b=c cosα or c=b/cosα.
Both legs → tangent: a=b tanα or b=a/tanα.

If both legs are known, the angle is α=arctan(a/b). A calculator’s tan⁻¹ usually means this inverse function, not 1/tanα. Set DEG for degree inputs. The other acute angle is 90°−α.

Some exact values make useful checks:

30°: sin30°=1/2, cos30°=√3/2, tan30°=1/√3.
45°: sine and cosine are √2/2; tangent is 1.
60°: sine is √3/2, cosine is 1/2, tangent is √3.

The 30° and 60° values follow from half an equilateral triangle; 45° comes from an isosceles right triangle.

Dividing a²+b²=c² by c² also gives sin²α+cos²α=1, while the ratios give tanα=sinα/cosα. These are consistency checks, not substitutes for identifying the correct sides.

Height: include the instrument level

An instrument is e=1.6 m above the ground. The horizontal distance to a vertical post is d=12 m, and the line of sight rises at α=35° from the horizontal. If the ground at the instrument and post base is at the same level:

$H=e+d\tan\alpha=1.6+12\tan(35\text{°})\approx10.0\ \text{m}.$

Tangent gives the height difference above the instrument, so add e. A higher or lower base requires a separate level correction. The value 10.0 is the rounded result of an ideal model, not a promise of field accuracy.

Sloping distance L along the line of sight is different. If that were known instead, the height difference would be L sinα and the horizontal projection L cosα. Do not substitute L for d in the tangent formula.

A 20% grade is not 20°

For a straight incline, rise h, horizontal run d and sloping length L form a right triangle. The percentage grade is:

$i=\frac{h}{d}\cdot100\%=\tan\alpha\cdot100\%.$

A rise of 0.6 m over 3 horizontal metres gives 20%. The angle is arctan(0.2)≈11.31°, and the sloping length is √(3²+0.6²)≈3.06 m. The denominator is therefore 3 m, rather than 3.06 m. At 45° the grade is 100%; it has no natural ceiling of 100 as a fraction of a whole would. Esri: three ways to express slope.

Acute angles and right triangles suffice here. Full-circle definitions, signs and periodicity belong to the algebra course’s trigonometry.