Math Vale

🏔️ Trigonometry Peaks · Trigonometry

Right Triangle Trigonometry

Use sine, cosine and tangent as ratios of sides in a right triangle to find unknown sides and angles.

In short

  • For a fixed angle the ratio of two sides is the same in every similar right triangle — that is why trigonometry works.
  • SOH-CAH-TOA: sine is opposite over hypotenuse, cosine adjacent over hypotenuse, tangent opposite over adjacent.
  • Label the sides relative to your angle before choosing a ratio; the hypotenuse always faces the right angle.
  • Two sides give an angle through an inverse ratio; an angle and a side give the missing side.

Why the ratios work at all

Take a right triangle with a 30° angle. Draw another one, twice as big, also with a 30° angle. The two triangles are similar, so all their matching sides are in the same ratio — which means the ratio of any two sides inside one triangle is exactly the same as in the other.

That is the whole idea of trigonometry. For a given acute angle, the ratio of two named sides depends only on the angle, never on the size of the triangle. So those ratios can be tabulated once and used forever.

The three sides are named relative to the angle you are working with:

  • hypotenuse — always the side facing the right angle, and always the longest
  • opposite — the side facing your angle
  • adjacent — the remaining side, next to your angle

The hypotenuse never changes, but "opposite" and "adjacent" swap over if you switch to the other acute angle. Label them before choosing a ratio.

SOH-CAH-TOA

The three ratios are

sin(A) = opposite / hypotenuse cos(A) = adjacent / hypotenuse tan(A) = opposite / adjacent

remembered as SOH-CAH-TOA: Sine-Opposite-Hypotenuse, Cosine-Adjacent-Hypotenuse, Tangent-Opposite-Adjacent.

In a 3-4-5 triangle, with the angle A opposite the side of length 3:

sin(A) = 3/5, cos(A) = 4/5, tan(A) = 3/4

Because the hypotenuse is the longest side, sine and cosine of an acute angle are always between 0 and 1. If you ever compute a sine of 2, something has gone wrong. Tangent has no such limit — it can be any positive number.

Finding a missing side

The routine is always the same:

1. Label opposite, adjacent and hypotenuse relative to the given angle. 2. Note which two of them appear in the problem (one known, one wanted). 3. Pick the ratio that uses exactly those two. 4. Write the equation and solve.

Angle 35°, hypotenuse 12, want the opposite side:

sin(35°) = a / 12 a = 12 x sin(35°) = 6.9 (to 1 decimal place)

If the unknown is on the bottom of the fraction instead, you divide rather than multiply:

tan(40°) = 9 / b -> b = 9 / tan(40°)

Two practical points. Keep the calculator in degree mode, and round only at the very end — rounding halfway through introduces error that compounds.

Finding a missing angle, and real situations

If two sides are known and the angle is wanted, form the ratio and then apply the inverse function, written sin-1, cos-1 or tan-1 (the shift key above sin, cos and tan).

opposite 5, adjacent 12 tan(A) = 5/12 = 0.4167 A = tan-1(0.4167) = 22.6° = 23° to the nearest degree

The inverse turns a ratio back into an angle; forgetting it and writing 0.4167 as the answer is a classic slip.

Two phrases appear constantly in word problems. The angle of elevation is measured up from the horizontal (looking up at a tower); the angle of depression is measured down from the horizontal (looking down from a cliff). Both are measured from a horizontal line, never from a vertical one.

The routine for a word problem is: sketch the triangle, mark the right angle, label the known sides and the angle, and only then choose the ratio.

Worked examples

Example 1

A right triangle has hypotenuse 20 cm and an angle of 25°. Find the side adjacent to that angle, to 1 decimal place.

  1. The known side is the hypotenuse and the wanted side is the adjacent, so use cosine (CAH).
  2. cos(25°) = b / 20.
  3. Multiply both sides by 20: b = 20 x cos(25°).
  4. cos(25°) = 0.9063, so b = 18.1 cm.

Example 2

A ramp is 15 m long and rises 4 m. Find the angle it makes with the ground, to the nearest degree.

  1. The ramp is the slanted side, so it is the hypotenuse; the 4 m rise is opposite the angle.
  2. Opposite with hypotenuse means sine: sin(A) = 4/15 = 0.2667.
  3. Apply the inverse: A = sin-1(0.2667).
  4. A = 15.5°, which is 15° to the nearest degree.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

In right triangle ABC the right angle is at C. Side a = 6 (opposite A), side b = 8 (opposite B) and the hypotenuse c = 10. Find sin(A). Give your answer as a fraction in lowest terms.

Answer: 3/5

  1. Relative to angle A: opposite = 6, adjacent = 8, hypotenuse = 10.
  2. sin(A) = opp/hyp
  3. sin(A) = 6/10

Problem 2

Difficulty 3 of 5

In right triangle ABC the right angle is at C, angle A = 70° and the hypotenuse c = 6 cm. How long is side b (adjacent to A)? Round to 1 decimal place.

Answer: 2.1 cm

  1. cos(70°) = adjacent / hypotenuse = b / 6
  2. b = 6 x cos(70°)
  3. cos(70°) = 0.342
  4. b = 2.1 cm

Problem 3

Difficulty 4 of 5

In right triangle ABC the right angle is at C. Side a = 9 (opposite A), side b = 12 (adjacent to A) and the hypotenuse c = 15. Find angle A, to the nearest degree.

Answer: 37 degrees

  1. tan(A) = 9/12 = 0.75
  2. A = tan-1(0.75)
  3. A = 36.87° = 37° to the nearest degree

Common mistakes

  • Choosing the wrong ratio because opposite and adjacent were not labelled first.
  • Writing the ratio upside down, so a sine comes out greater than 1.
  • Giving the ratio as the answer instead of applying sin-1, cos-1 or tan-1.
  • Leaving the calculator in radian mode, or rounding partway through the calculation.

What you should be able to do

  • Identify the opposite, adjacent and hypotenuse sides for an angle.
  • Use SOH-CAH-TOA to find a missing side.
  • Use inverse trig functions to find a missing angle.
  • Solve angle-of-elevation and depression problems.

Where this fits in the curriculum

Common Core

  • HSG-SRT.C.6

    High school — Understand that side ratios in right triangles are properties of the angles, leading to definitions of sine, cosine and tangent.

  • HSG-SRT.C.7

    High school — Explain and use the relationship between the sine and cosine of complementary angles.

  • HSG-SRT.C.8

    High school — Use trigonometric ratios and the Pythagorean theorem to solve right triangles in applied problems.

SAT

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