Unit 5, lesson 50

Sine, Cosine and Tangent in Right Triangles

How to name the opposite and adjacent sides for a chosen acute angle, why sine, cosine and tangent depend only on the angle, and how to use them and complementary angles to find sides.

  • Theory part 99
  • Practice part 100
  • Right Triangle Trigonometry and the Unit Circle

Theory, part 99

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Practice, part 100

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A wheelchair ramp may rise at most a few degrees, and a builder knows the angle long before knowing the lengths. The question is how an angle can tell you anything about lengths. The answer is that every right triangle with the same acute angle has the same shape, so the ratios of its sides are fixed by that angle. Those ratios have names, sine, cosine and tangent, and once you can pick the right one you can turn an angle and one side into any other side.

See also Right Triangle Trigonometry in Precalculus.

Key terms

Opposite, adjacent and hypotenuse

Definition Choose one acute angle θ\theta of a right triangle as the reference angle. The hypotenuse is the side opposite the right angle. The opposite side is the leg that does not touch the vertex of θ\theta. The adjacent side is the leg that does touch the vertex of θ\theta. The labels opposite and adjacent depend on which acute angle you chose, while the hypotenuse is the same for both.

In plain words Stand at the corner with angle θ\theta. The leg you are touching is adjacent, the leg across from you is opposite, and the long side is the hypotenuse no matter where you stand.

Example Triangle ABCABC has its right angle at CC, with BC=3BC = 3, AC=4AC = 4 and AB=5AB = 5.

Reference angleOppositeAdjacentHypotenuse
AABC=3BC = 3AC=4AC = 4AB=5AB = 5
BBAC=4AC = 4BC=3BC = 3AB=5AB = 5

Switching the reference angle from AA to BB swaps opposite and adjacent and leaves the hypotenuse alone.

Sine

Definition For an acute angle θ\theta of a right triangle,

sin⁡θ=oppositehypotenuse.\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}.

The notation sin⁡θ\sin \theta is read “sine of theta”. Because a leg is shorter than the hypotenuse, 0<sin⁡θ<10 < \sin \theta < 1 for every acute angle.

In plain words Sine tells you what fraction of the hypotenuse the opposite leg is.

Example In the 3, 4, 5 triangle above,

sin⁡A=35,sin⁡B=45.\sin A = \frac{3}{5}, \qquad \sin B = \frac{4}{5}.

Cosine

Definition For an acute angle θ\theta of a right triangle,

cos⁡θ=adjacenthypotenuse.\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}.

The notation cos⁡θ\cos \theta is read “cosine of theta”, and 0<cos⁡θ<10 < \cos \theta < 1 for every acute angle.

In plain words Cosine tells you what fraction of the hypotenuse the adjacent leg is.

Example In the same triangle,

cos⁡A=45,cos⁡B=35.\cos A = \frac{4}{5}, \qquad \cos B = \frac{3}{5}.

Notice that cos⁡A\cos A equals sin⁡B\sin B. That is no accident, and the section on complementary angles explains it.

Tangent

Definition For an acute angle θ\theta of a right triangle,

tan⁡θ=oppositeadjacent=sin⁡θcos⁡θ.\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin \theta}{\cos \theta}.

The notation tan⁡θ\tan \theta is read “tangent of theta”. Tangent can be any positive number, since nothing forces the opposite leg to be shorter than the adjacent one.

In plain words Tangent compares the two legs. It is the rise over the run of the hypotenuse when the adjacent leg lies flat.

Example In the 3, 4, 5 triangle,

tan⁡A=34,tan⁡B=43.\tan A = \frac{3}{4}, \qquad \tan B = \frac{4}{3}.

The SAT reference sheet does not list these three definitions, so you need to know them yourself. A common memory aid is SOH CAH TOA, for sine is opposite over hypotenuse, cosine is adjacent over hypotenuse and tangent is opposite over adjacent.

Similarity and the ratios

Definition Any two right triangles with an acute angle of the same measure θ\theta are similar by AA, since they share the right angle and the angle θ\theta. Corresponding sides of similar triangles are proportional, so the ratios opposite over hypotenuse, adjacent over hypotenuse and opposite over adjacent are the same in both triangles. That is why sin⁡θ\sin \theta, cos⁡θ\cos \theta and tan⁡θ\tan \theta depend only on the angle θ\theta and not on the size of the triangle.

In plain words Blow up or shrink a right triangle and its angles stay the same. Every side is multiplied by the same factor, so every ratio of two sides stays the same too.

Example Doubling the 3, 4, 5 triangle gives a triangle with sides 66, 88 and 1010 and the same angle AA. Then

sin⁡A=610=35,\sin A = \frac{6}{10} = \frac{3}{5},

exactly as before. This also works backwards. If a question says sin⁡θ=513\sin \theta = \frac{5}{13}, you may draw any right triangle with opposite 55 and hypotenuse 1313. Its adjacent leg is 169−25=12\sqrt{169 - 25} = 12, so

cos⁡θ=1213,tan⁡θ=512.\cos \theta = \frac{12}{13}, \qquad \tan \theta = \frac{5}{12}.

Finding a missing side

Definition If you know an acute angle θ\theta and one side of a right triangle, choose the ratio that contains the known side and the unknown side, set it equal to sin⁡θ\sin \theta, cos⁡θ\cos \theta or tan⁡θ\tan \theta, and solve. A calculator gives decimal values of these ratios, and its angle mode must match the unit of the angle. The angles in this lesson are in degrees, so the calculator must be in degree mode. The Desmos graphing calculator built into the testing app has a setting to switch between radians and degrees.

In plain words Name the sides from the given angle, pick the ratio that links what you have with what you want, then solve a one step equation.

Example A right triangle has hypotenuse 2020 and an acute angle of 35∘35^\circ. The side opposite that angle is xx. Opposite and hypotenuse point to sine, so

sin⁡35∘=x20,x=20sin⁡35∘≈11.47.\begin{gathered} \sin 35^\circ = \frac{x}{20}, \\[0.4em] x = 20 \sin 35^\circ \approx 11.47. \end{gathered}

In radian mode the same keystrokes compute 20sin⁡(35)20 \sin(35) with 3535 radians, about −8.56-8.56, which is not even positive. A negative or otherwise impossible length is a sign that the mode is wrong.

Complementary angles

Definition Two angles are complementary when their measures add up to 90∘90^\circ. The two acute angles of a right triangle are complementary, and the side opposite one of them is adjacent to the other. So for every acute angle θ\theta, measured in degrees,

sin⁡θ=cos⁡(90∘−θ),cos⁡θ=sin⁡(90∘−θ).\begin{gathered} \sin \theta = \cos(90^\circ - \theta), \\[0.4em] \cos \theta = \sin(90^\circ - \theta). \end{gathered}

For acute angles α\alpha and β\beta the equation sin⁡α=cos⁡β\sin \alpha = \cos \beta holds exactly when α+β=90∘\alpha + \beta = 90^\circ.

In plain words The co in cosine stands for complement. The sine of one acute angle in a right triangle is the cosine of the other one.

Example Since 32+58=9032 + 58 = 90, the value sin⁡32∘\sin 32^\circ equals cos⁡58∘\cos 58^\circ, and both are about 0.5300.530. A typical SAT question gives sin⁡((2x)∘)=cos⁡((x+15)∘)\sin((2x)^\circ) = \cos((x + 15)^\circ) for acute angles and asks for xx. The angles must be complementary, so

2x+(x+15)=90,x=25.2x + (x + 15) = 90, \qquad x = 25.

Check that both angles are acute, 50∘50^\circ and 40∘40^\circ.

Check yourself

  1. In right triangle ABCABC the right angle is at CC, AC=12AC = 12, BC=5BC = 5 and AB=13AB = 13. What is the value of cos⁡B\cos B? (A) 512\frac{5}{12} (B) 513\frac{5}{13} (C) 1213\frac{12}{13} (D) 125\frac{12}{5}
Show the answer

The correct choice is (B). From angle BB the adjacent leg is BC=5BC = 5 and the hypotenuse is AB=13AB = 13, so cos⁡B=513\cos B = \frac{5}{13}. Choice (C) is cos⁡A\cos A, which uses the adjacent side of the wrong angle. Choices (A) and (D) are tan⁡A\tan A and tan⁡B\tan B, ratios of the two legs.

  1. For acute angles, sin⁡(x∘)=cos⁡((3x−10)∘)\sin(x^\circ) = \cos((3x - 10)^\circ). What is the value of xx?
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The answer is 2525. Sine of one acute angle equals cosine of another exactly when the angles are complementary, so x+3x−10=90x + 3x - 10 = 90, which gives 4x=1004x = 100 and x=25x = 25. The angles are 25∘25^\circ and 65∘65^\circ, both acute. Setting the angles equal instead would give x=5x = 5, but sin⁡5∘\sin 5^\circ is not cos⁡5∘\cos 5^\circ.

  1. A ladder 44 meters long leans against a vertical wall and makes an angle of 70∘70^\circ with the level ground. To the nearest tenth of a meter, how high up the wall does the ladder reach? (A) 1.41.4 (B) 3.13.1 (C) 3.83.8 (D) 11.011.0
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The correct choice is (C). The height is opposite the 70∘70^\circ angle and the ladder is the hypotenuse, so the height is 4sin⁡70∘≈3.764 \sin 70^\circ \approx 3.76, which rounds to 3.83.8. Choice (A) uses cosine and gives the distance along the ground. Choice (D) uses tangent, and a height of 1111 meters is impossible for a 44 meter ladder. Choice (B) is what a calculator in radian mode gives for 4sin⁡704 \sin 70.

How this lesson connects