Exact Trigonometric Ratios with Surds

An exact side from an exact ratio

A right triangle has a hypotenuse of 8 and an angle of 60°. The side opposite the angle is 8 sin 60°. A calculator gives sin 60° as 0.8660…, a decimal that never ends, so 8 sin 60° from a calculator is already rounded. But $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2 exactly, and that gives the side exactly.

Multiply: $8\times \frac{\sqrt{3}}{2}=\left(8÷2\right)\times \sqrt{3}=4\sqrt{3}$8 × √3/2 = (8 ÷ 2) × √3 = 4√3. The 2 in the denominator divides into the 8, and the $\sqrt{3}$√3 is left as it is.

The side next to the angle is 8 cos 60° = 8 × ½ = 4. So the triangle has sides 4, $4\sqrt{3}$4√3 and 8, all exact.

Check by Pythagoras. ${\left(4\sqrt{3}\right)}^2=4^2\times {\left(\sqrt{3}\right)}^2=16\times 3=48$(4√3)² = 4² × (√3)² = 16 × 3 = 48, and $48+4^2=48+16=64=8^2$48 + 4² = 48 + 16 = 64 = 8². The two exact sides fit the hypotenuse exactly.

The hypotenuse is 8 and the angle is 60°. The opposite side is $8\times \frac{\sqrt{3}}{2}=4\sqrt{3}$8 × √3/2 = 4√3, and the adjacent side is 8 × ½ = 4.

The same at 30° and 45°

With the same hypotenuse of 8 and an angle of 30°, the opposite side is 8 sin 30° = 8 × ½ = 4, and the adjacent side is $8\cos 30^{\circ }=8\times \frac{\sqrt{3}}{2}=4\sqrt{3}$8 cos 30° = 8 × √3/2 = 4√3. The 30° angle faces the shortest side, so the two legs have swapped places.

At 45° the two legs are equal. With a hypotenuse of 10, each one is $10\sin 45^{\circ }=10\times \frac{\sqrt{2}}{2}=5\sqrt{2}$10 sin 45° = 10 × √2/2 = 5√2. Check: ${\left(5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2=50+50=100=10^2$(5√2)² + (5√2)² = 50 + 50 = 100 = 10².

A surd in the denominator

Now the angle is 30°, the side next to it is 5, and the height opposite it is wanted. The tangent pairs those two sides: the height is 5 tan 30°. The exact value is $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3, so the height is $5\times \frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}$5 × 1/√3 = 5/√3.

That is exact, but $\sqrt{3}$√3 is still in the denominator. The usual form has a whole number in the denominator. To get it, rationalize: multiply the numerator and the denominator by $\sqrt{3}$√3. Since $\sqrt{3}\times \sqrt{3}=3$√3 × √3 = 3, $\frac{5}{\sqrt{3}}=\frac{\left(5\times \sqrt{3}\right)}{\left(\sqrt{3}\times \sqrt{3}\right)}=\frac{5\sqrt{3}}{3}$5/√3 = (5 × √3)/(√3 × √3) = 5√3/3.

Multiplying the top and the bottom by the same number is multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$√3/√3, which is 1, so the value does not change. Only its form does. In the rationalized form the size is easier to see: $\sqrt{3}$√3 is about 1.732, so $\frac{5\sqrt{3}}{3}$5√3/3 is about 8.66 ÷ 3 = 2.89.

The angle is 30° and the adjacent side is 5. The height is $5\tan 30^{\circ }=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}$5 tan 30° = 5/√3 = 5√3/3, and the hypotenuse is $\frac{10\sqrt{3}}{3}$10√3/3.

Finishing the triangle

The hypotenuse of the same triangle is $5/\cos 30^{\circ }=5÷\frac{\sqrt{3}}{2}=5\times \frac{2}{\sqrt{3}}=\frac{10}{\sqrt{3}}$5 / cos 30° = 5 ÷ √3/2 = 5 × 2/√3 = 10/√3. Rationalize again: $\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}$10/√3 = 10√3/3.

Check by Pythagoras. ${\left(\frac{5\sqrt{3}}{3}\right)}^2=\frac{\left(25\times 3\right)}{9}=\frac{75}{9}=\frac{25}{3}$(5√3/3)² = (25 × 3)/9 = 75/9 = 25/3, and $\frac{25}{3}+5^2=\frac{25}{3}+\frac{75}{3}=\frac{100}{3}$25/3 + 5² = 25/3 + 75/3 = 100/3. On the other side, ${\left(\frac{10\sqrt{3}}{3}\right)}^2=\frac{\left(100\times 3\right)}{9}=\frac{300}{9}=\frac{100}{3}$(10√3/3)² = (100 × 3)/9 = 300/9 = 100/3. The two agree exactly.

Dividing by a ratio

Sometimes the side wanted is the one the ratio divides by. A right triangle has an angle of 60° with the side opposite it equal to 6. Since tan 60° = opposite ÷ adjacent, the adjacent side is $6/\tan 60^{\circ }=\frac{6}{\sqrt{3}}$6 / tan 60° = 6/√3.

Rationalize: $\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}$6/√3 = 6√3/3, and the 3 divides into the 6, so the side is $2\sqrt{3}$2√3. Rationalizing often lets a fraction simplify like this.

Why keep the surd

The exact side $4\sqrt{3}$4√3 is about 6.93, to 2 decimal places. Square both. ${\left(4\sqrt{3}\right)}^2=16\times 3=48$(4√3)² = 16 × 3 = 48 exactly, but $6.93^2=48.0249$6.93² = 48.0249. The rounded value is already a little wrong, and squaring it carries the error into the next answer.

The error grows when an early rounding is multiplied. A length is $50\sqrt{3}$50√3, which is 86.60 to 2 decimal places. Round $\sqrt{3}$√3 to 1.7 first, and 50 × 1.7 = 85.0, wrong by more than 1.5. With 1.73, 50 × 1.73 = 86.5, still wrong in the first decimal place. Only with 1.732 does 50 × 1.732 = 86.6 come out right to 1 decimal place.

So keep $\sqrt{3}$√3 and $\sqrt{2}$√2 exact through every step, and round once, at the end, to the accuracy the question asks for. An exact answer can always be rounded later; a rounded one cannot be made exact again.

The usual mistakes

Using the value for the other angle. sin 30° = ½ and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2. The 60° angle faces the longer leg, so with a hypotenuse of 8 the side opposite 60° is $4\sqrt{3}$4√3, not 4.

Leaving out the half. sin 60° is $\frac{\sqrt{3}}{2}$√3/2, not $\sqrt{3}$√3, so 8 sin 60° is $4\sqrt{3}$4√3, not $8\sqrt{3}$8√3. A leg of $8\sqrt{3}$8√3 would be longer than the hypotenuse of 8.

Multiplying only the top by $\sqrt{3}$√3. That changes the value: $\frac{5}{\sqrt{3}}$5/√3 is about 2.89, but $\frac{5\sqrt{3}}{\sqrt{3}}=5$5√3/√3 = 5. Multiply the numerator and the denominator together.

Rounding $\sqrt{3}$√3 early. Write 1.732… only on the last line, once the exact answer is found.

Two sightings of a cliff

In the application below, a cliff is sighted at angles of elevation of 30° and 60°, measured up from the level ground. Each sighting makes a right triangle, and the height of the cliff is the opposite side in both. Writing $\tan 60^{\circ }=\sqrt{3}$tan 60° = √3 and $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3 exactly lets the two triangles be solved together, and the height comes out as an exact surd before it is rounded once.