Year 11 Extended · Unit 1 Right-angled triangles and circle geometry · Lesson 3 of 7

Exact values: why 30°, 45° and 60° are special, and what else they give

A set square has angles that are worth knowing exactly. This lesson derives their sines, cosines and tangents from two triangles, uses them to prove statements with no calculator, and then builds one new exact value, tan 15°, from them.

Geometry and TrigonometryNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 10: simplify √48.
    Answer4√3, because 48 = 16 × 3.
  2. From Year 10: write 10 ÷ √5 with a whole-number denominator.
    Answer2√5, because 10√5 ÷ 5 = 2√5.
  3. From Year 9: the legs of a right-angled triangle are both 7 cm. Find the hypotenuse exactly.
    Answer7√2 cm.
Words used in this lesson
exact value
a number written with its roots and fractions, not rounded. For example: √3/2, not 0.866.
surd
a root that does not come out as a whole number or fraction. For example: √5.
rationalize
rewrite a fraction so the bottom has no root. For example: 1/√2 becomes √2/2.
isosceles
a triangle with two equal sides and two equal angles. For example: a triangle with two sides of 5 cm.
exterior angle
the angle outside a triangle formed by extending a side. For example: equals the sum of the two opposite interior angles.

A set square and a ruler

A set square has angles of 30°, 60° and 90°, and its shortest side is exactly 10 cm. Nobody has a calculator in this task.

Predict first: is the longest side a whole number of centimetres? Is the third side? Give your reasons, then use the first figure with a hypotenuse of 20 and an angle of 30° to test them.

60
12
6√3 cmopposite side
6 cmadjacent side

Hypotenuse 12, angle 60°: opposite 12 × sin 60° = 6√3 cm, adjacent 12 × cos 60° = 6 cm.

Show me the steps
  1. Hypotenuse 12, angle 60°: opposite is 12 × sin 60° = 6√3 and adjacent is 12 × cos 60° = 6. The sides are in the ratio √3 : 1 : 2.
  2. Keep 12 and set the angle to 30°: the two short sides swap, opposite 6 and adjacent 6√3. The same triangle, seen from the other acute angle.
  3. Set the angle to 45° with hypotenuse 12: both short sides are 12 × sin 45° = 6√2. The sides are in the ratio 1 : 1 : √2.
  4. Change the hypotenuse to 20 and the angle to 30°: opposite 10, adjacent 10√3. Whatever the size, the coefficient of the surd is half the hypotenuse.

Half an equilateral triangle gives 30° and 60°; half a square gives 45°. Cut an equilateral triangle of side 2k down the middle: the halves have sides k, k√3 and 2k. Cut a square of side k along a diagonal: the halves have sides k, k and k√2.

So sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3; sin 45° = cos 45° = 1/√2 and tan 45° = 1; sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3. The ratio is the same at every size, because every triangle with these angles is similar.

3
2 − √3tan 15° exactly
11.2 cmlength BD

k = 3: BD = 6 + 3√3 ≈ 11.2 cm, and tan 15° = 3 ÷ (6 + 3√3) = 2 − √3 ≈ 0.268.

Show me the steps
  1. k = 3: AB = 3, AC = 6, BC = 3√3. Extending to D with CD = 6 gives BD = 6 + 3√3 ≈ 11.2. In the triangle ABD, tan D = 3 ÷ (6 + 3√3).
  2. Rationalizing: 1 ÷ (2 + √3) = (2 − √3) ÷ ((2 + √3)(2 − √3)) = (2 − √3) ÷ 1. So tan 15° = 2 − √3, and k has cancelled.
  3. Change k to 5: AB = 5, BD = 10 + 5√3 ≈ 18.7, and the ratio AB ÷ BD is still 2 − √3 ≈ 0.268. The size of the construction does not change the angle.
  4. The angle D is half of 30° because the triangle ACD is isosceles with exterior angle 30°. This is the move that makes a new exact value out of a known one.

Exact is not “nearest decimal”. √3 ÷ 2 is not 0.87 and 1 ÷ √2 is not 0.71: they are numbers whose decimals never end. In a proof, write √3, not 1.73, because once you round, equality is lost and “show that” cannot be shown.

Worked examples

1. An equilateral triangle has side 10 cm. Find its area exactly.

  1. Half the triangle has hypotenuse 10 and short side 5, so the height is 5√3.
  2. Area = ½ × 10 × 5√3 = 25√3 cm².

2. Find the exact value of (tan 60° − tan 30°) ÷ (1 + tan 60° tan 30°).

  1. tan 60° = √3 and tan 30° = 1/√3 = √3/3.
  2. Top: √3 − √3/3 = 2√3/3. Bottom: 1 + √3 × 1/√3 = 2.
  3. The quotient is (2√3/3) ÷ 2 = √3/3, which is tan 30°.

3. Prove that tan 15° = 2 − √3.

  1. Take a triangle ABC with a right angle at B and an angle of 30° at C, with AB = 1. Then AC = 2 and BC = √3.
  2. Extend BC to D with CD = 2 = CA. Triangle ACD is isosceles and its exterior angle at C is 30°, so each base angle, including angle D, is 15°.
  3. In the right-angled triangle ABD, BD = 2 + √3, so tan 15° = AB ÷ BD = 1 ÷ (2 + √3).
  4. Multiply top and bottom by (2 − √3): the bottom is 4 − 3 = 1, so tan 15° = 2 − √3.

Your go. The first step is given. Find the exact value of sin 60° ÷ tan 60°.

  1. sin 60° = √3/2 and tan 60° = √3.
  2. Write the quotient as a fraction and cancel the √3.
Check the answer

(√3/2) ÷ √3 = 1/2. Because sin ÷ tan = cos, the value is cos 60°, and cos 60° is exactly 1/2. Writing 0.866 ÷ 1.732 would agree to the digits shown but would not prove it.

Check before you go on

1. A right-angled triangle has angles of 30°, 60° and 90°. The side opposite the 60° angle is 9 cm. What is the exact length of the side opposite the 30° angle?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A right-angled triangle has angles of 30°, 60° and 90°, and its hypotenuse is 10 cm. Its area is ?√3 cm². What number goes in front of the √3, to 1 decimal place?

3. Without a calculator, sin 45° × tan 60° can be written as √k ÷ 2. What is k?

Level 2 · Building

4. In triangle ABC, the angle at B is 90°, the angle at A is 60° and AC is 16 cm. Find the perimeter, to 1 decimal place, using √3 ≈ 1.732 at the last step.

5. The sun is 15° above the horizon and a mast is 30 m tall. Using tan 15° = 2 − √3, find the length of its shadow, to the nearest metre.

6. An acute angle x satisfies √3 tan x = 1. Find x in degrees.

Level 3 · Stretch

7. A regular hexagon has sides of 6 cm. Its area is ?√3 cm². What number goes in front of the √3?

8. In triangle ABC the angle at B is 90°, the angle at C is 30° and AB = 4 cm. The point D is on BC with the angle ADB equal to 45°. Find the length DC, to 2 decimal places, using √3 ≈ 1.732 at the last step.

Something to think about

Test the idea that sin 60° is twice sin 30°, using the exact values. Is it true? Then try sin 2x against 2 sin x for x = 30° and for x = 45°, and say what the pattern of your results tells you.

One way to think about it

Not true: sin 60° = √3 ÷ 2 ≈ 0.866 and 2 sin 30° = 1. For x = 45°, sin 90° = 1 but 2 sin 45° = √2 ≈ 1.41. Doubling an angle does not double its sine. The claim fails for every acute angle, not only large ones: at 30° the doubled value is exactly 1 but sin 60° is less than 1, and at 45° the doubled value √2 is more than any sine can be.

Exit ticket

9. A right-angled triangle has angles of 45°, 45° and 90°, and its hypotenuse is 10 cm. What is the exact length of each of its two equal sides?