Unit 4, lesson 50

Special Right Triangles

The exact side ratios of the 45-45-90 and 30-60-90 triangles, how they give exact trigonometric values at 30, 45 and 60 degrees, and how to rationalize a denominator.

  • Theory part 99
  • Practice part 100
  • Triangles and the Unit Circle

Cut a square along its diagonal, or cut an equilateral triangle down the middle, and you get right triangles whose sides you can find exactly. Their angles of 30∘30^\circ, 45∘45^\circ and 60∘60^\circ come up again and again in trigonometry and calculus. Knowing their ratios as exact numbers, rather than calculator decimals, keeps your answers exact and saves time.

Key terms

The 45-45-90 triangle

Definition A 45-45-90 triangle is a right triangle whose two acute angles are both 45∘45^\circ, or π4\frac{\pi}{4} radians. It is isosceles, and its sides are in the ratio

leg:leg:hypotenuse=1:1:2.\text{leg} : \text{leg} : \text{hypotenuse} = 1 : 1 : \sqrt{2}.

To see why, cut a square of side s>0s > 0 along a diagonal. Each half is a right triangle with legs ss and ss. Its two acute angles are equal, since they face equal sides, and they add up to 90∘90^\circ, so each is 45∘45^\circ. By the Pythagorean theorem its hypotenuse is s2+s2=s2\sqrt{s^2 + s^2} = s\sqrt{2}. Every 45-45-90 triangle is similar to this one, so every one has these ratios.

In plain words This triangle is half a square. The two legs are equal, and the hypotenuse is a leg times 2\sqrt{2}, about 1.4141.414 times as long.

Example A square tile with side 7 centimeters has a diagonal of 72≈9.907\sqrt{2} \approx 9.90 centimeters. Going the other way, a 45-45-90 triangle with hypotenuse 10 has legs of length

102=1022=52≈7.07.\frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07.

The middle step multiplies the top and bottom by 2\sqrt{2}, which is explained under rationalizing denominators below.

The 30-60-90 triangle

Definition A 30-60-90 triangle is a right triangle with acute angles 30∘30^\circ and 60∘60^\circ, or π6\frac{\pi}{6} and π3\frac{\pi}{3} radians. The short leg is opposite the 30∘30^\circ angle, the long leg is opposite the 60∘60^\circ angle, and the sides are in the ratio

short leg:long leg:hypotenuse=1:3:2.\begin{aligned} &\text{short leg} : \text{long leg} : \text{hypotenuse} \\[0.4em] &\quad = 1 : \sqrt{3} : 2. \end{aligned}

To see why, take an equilateral triangle with side 2 and draw the segment from one vertex to the midpoint of the opposite side. By symmetry this segment is perpendicular to that side, so it splits the triangle into two right triangles with angles 30∘30^\circ at the top and 60∘60^\circ at the base, hypotenuse 2 and short leg 1. The long leg is 22−12=3\sqrt{2^2 - 1^2} = \sqrt{3}. Every 30-60-90 triangle is similar to this one, so its sides are ss, s3s\sqrt{3} and 2s2s for some s>0s > 0.

In plain words This triangle is half an equilateral triangle. The hypotenuse is twice the short leg, and the long leg is the short leg times 3\sqrt{3}.

Example A 30-60-90 triangle with hypotenuse 12 has short leg 6 and long leg 63≈10.396\sqrt{3} \approx 10.39. If instead the long leg is 9, the short leg is

93=933=33,\frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3},

and the hypotenuse is twice that, 636\sqrt{3}.

Exact ratios

Definition An exact value is a number written without rounding, using integers, fractions, radicals such as 3\sqrt{3} and constants such as π\pi. A decimal such as 0.8660.866 is only an approximation of 32\frac{\sqrt{3}}{2}, not equal to it. The special triangles give exact values for the trigonometric ratios of 30∘30^\circ, 45∘45^\circ and 60∘60^\circ. For example, the 30-60-90 triangle with sides 1, 3\sqrt{3} and 2 gives

sin⁡30∘=12,cos⁡30∘=32,tan⁡30∘=13.\begin{gathered} \sin 30^\circ = \frac{1}{2}, \qquad \cos 30^\circ = \frac{\sqrt{3}}{2}, \\[0.8em] \tan 30^\circ = \frac{1}{\sqrt{3}}. \end{gathered}

In plain words Keep the square roots. A rounded decimal is close but not equal, and the small error can grow as you keep computing.

Example The exact values for 60∘60^\circ give

sin⁡260∘+cos⁡260∘=(32)2+(12)2=34+14=1.\begin{aligned} &\sin^2 60^\circ + \cos^2 60^\circ \\[0.4em] &\quad = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \\[0.8em] &\quad = \frac{3}{4} + \frac{1}{4} = 1. \end{aligned}

With the rounded value 0.8660.866 in place of 32\frac{\sqrt{3}}{2}, the same sum is 0.749956+0.25=0.9999560.749956 + 0.25 = 0.999956, close to 1 but not equal to it.

Rationalizing denominators

Definition To rationalize a denominator is to rewrite a fraction as an equal fraction with no radical in the denominator, by multiplying the numerator and the denominator by the same nonzero expression. For b>0b > 0,

ab=ab⋅bb=abb.\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}.

For a denominator p+bp + \sqrt{b} or p−bp - \sqrt{b}, with rational numbers pp and b>0b > 0 and p2≠bp^2 \ne b, multiply the numerator and the denominator by the conjugate, which is the same expression with the other sign. The condition p2≠bp^2 \ne b makes sure neither expression is zero, and their product is rational,

(p+b)(p−b)=p2−b.(p + \sqrt{b})(p - \sqrt{b}) = p^2 - b.

In plain words You multiply by a fraction equal to 1, chosen so that the square root leaves the denominator. The value does not change. Both 13\frac{1}{\sqrt{3}} and 33\frac{\sqrt{3}}{3} are correct, and the rationalized form is the usual way to report an answer because it is easier to compare and to add.

Example With a single square root in the denominator, multiply by that root.

63=633=23.\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}.

With a sum or difference in the denominator, use the conjugate.

12−3=2+3(2−3)(2+3)=2+34−3=2+3.\begin{aligned} \frac{1}{2 - \sqrt{3}} &= \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} \\[0.8em] &= \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}. \end{aligned}

The six ratios of the special angles

Definition For θ=30∘,45∘,60∘\theta = 30^\circ, 45^\circ, 60^\circ, which are π6,π4,π3\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} in radians, the six trigonometric ratios have these exact values, read from the triangles with sides 1, 3\sqrt{3}, 2 and 1, 1, 2\sqrt{2} and written with rationalized denominators.

θ\thetasin⁡θ\sin \thetacos⁡θ\cos \thetatan⁡θ\tan \thetacsc⁡θ\csc \thetasec⁡θ\sec \thetacot⁡θ\cot \theta
30∘=π630^\circ = \frac{\pi}{6}12\frac{1}{2}32\frac{\sqrt{3}}{2}33\frac{\sqrt{3}}{3}22233\frac{2\sqrt{3}}{3}3\sqrt{3}
45∘=π445^\circ = \frac{\pi}{4}22\frac{\sqrt{2}}{2}22\frac{\sqrt{2}}{2}112\sqrt{2}2\sqrt{2}11
60∘=π360^\circ = \frac{\pi}{3}32\frac{\sqrt{3}}{2}12\frac{1}{2}3\sqrt{3}233\frac{2\sqrt{3}}{3}2233\frac{\sqrt{3}}{3}

In plain words There is no need to memorize 18 numbers. Sketch the two triangles, label the sides, and read off any ratio. Notice that sine climbs through 12,22,32\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2} while cosine runs through the same values in reverse.

Example A ladder 8 meters long leans against a vertical wall at an angle of 60∘60^\circ with the level ground. The ladder is the hypotenuse, so it reaches a height of

8sin⁡60∘=8⋅32=43≈6.93 meters,\begin{aligned} 8 \sin 60^\circ &= 8 \cdot \frac{\sqrt{3}}{2} \\[0.8em] &= 4\sqrt{3} \approx 6.93 \text{ meters,} \end{aligned}

and its foot stands 8cos⁡60∘=48 \cos 60^\circ = 4 meters from the wall, exactly.

Complementary angles

Definition Two angles are complementary when they 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

sin⁡(90∘−θ)=cos⁡θ,tan⁡(90∘−θ)=cot⁡θ,sec⁡(90∘−θ)=csc⁡θ,\begin{aligned} \sin(90^\circ - \theta) &= \cos \theta, \\[0.4em] \tan(90^\circ - \theta) &= \cot \theta, \\[0.4em] \sec(90^\circ - \theta) &= \csc \theta, \end{aligned}

and the same holds with the two functions in each equation swapped. Sine and cosine are called cofunctions, and so are tangent and cotangent, and secant and cosecant.

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

Example In the 30-60-90 triangle the side of length 1 is opposite 30∘30^\circ and adjacent to 60∘60^\circ, so sin⁡30∘=cos⁡60∘=12\sin 30^\circ = \cos 60^\circ = \frac{1}{2}. In the same way tan⁡60∘=cot⁡30∘=3\tan 60^\circ = \cot 30^\circ = \sqrt{3}. The angle 45∘45^\circ is its own complement, which is why sin⁡45∘=cos⁡45∘\sin 45^\circ = \cos 45^\circ.

Check yourself

  1. A 45-45-90 triangle has hypotenuse 8. Find the exact length of each leg.
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The hypotenuse is a leg times 2\sqrt{2}, so each leg is

82=822=42.\frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}.
  1. In a 30-60-90 triangle the side opposite the 60∘60^\circ angle has length 15. Find the other two sides exactly.
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That side is the long leg, which is the short leg times 3\sqrt{3}. So the short leg is

153=1533=53,\frac{15}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3},

and the hypotenuse is twice the short leg, 10310\sqrt{3}.

  1. Find the exact value of tan⁡30∘+cot⁡30∘\tan 30^\circ + \cot 30^\circ with a rationalized denominator.
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From the table, tan⁡30∘=33\tan 30^\circ = \frac{\sqrt{3}}{3} and cot⁡30∘=3=333\cot 30^\circ = \sqrt{3} = \frac{3\sqrt{3}}{3}, so the sum is

33+333=433.\frac{\sqrt{3}}{3} + \frac{3\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}.

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