Using right-angled triangle ABN where hypotenuse AB = 2, base BN = 1, and perpendicular AN = √(3), what is the exact value of 30° given that BAN = 30°?

√(3)2. In right-angled triangle ABN, 30° = (adjacent)/(hypotenuse) = (AN)/(AB) = √(3)2.

Grade 10 · Core Mathematics · Lesson 7 of 11

Trigonometric Ratios of Special Angles: 30° and 60°

A Grade 10 Core Mathematics CBE lesson from the sub-strand Trigonometry 1. In it, learners derive exact trigonometric ratios for 30° and 60° using an equilateral triangle and its perpendicular bisector.

Strand Measurements and GeometrySub-strand Trigonometry 1Lesson code 10CM-24-L740 minutes · 3 slides

KICD learning outcome (d)

Determine trigonometric ratios of special angles: 30°, 45°, 60° and 90° using triangles.

By the end of this lesson, learners should be able to

Derive exact trigonometric ratios for 30° and 60° using an equilateral triangle and its perpendicular bisector.
Summarise sine, cosine, and tangent values for 30° and 60° in exact surd form.
Simplify trigonometric expressions and solve geometric problems involving 30° and 60° angles without tables or calculators.

Key vocabulary

Perpendicular bisector: A line segment dividing another segment into two equal halves at a right angle.; The perpendicular bisector $A N$ cuts side $B C$ of length 2 into two 1 unit lengths.
Surd form: An exact representation of an irrational number using a root sign without decimal approximation.; The altitude $A N$ equals $\sqrt{3}$ units, which is written in exact surd form.
Trigonometric ratio: A ratio comparing lengths of two sides of a right-angled triangle for a given angle.; The trigonometric ratio $\sin 30^{\circ}$ compares the opposite side to the hypotenuse.
Altitude: The perpendicular distance from a geometric vertex to the opposite base side.; The altitude of the equilateral triangle splits the base into two equal parts.

Lesson preview

Slide 1 · Deriving Trigonometric Ratios of 30° and 60°Opening · 12 min

Figure 1: An equilateral triangle is split in half by a perpendicular bisector to form two right-angled triangles with angles of 30 and 60 degrees.

We determine the exact trigonometric ratios of special angles $30^{\circ}$ and $60^{\circ}$ using an equilateral triangle $A B C$ with side length $2 units$ . Drawing the perpendicular bisector $A N$ to side $B C$ splits $A B C$ into two congruent right-angled triangles $A B N$ and $A C N$ . The bisector creates an angle of $30^{\circ}$ at $A$ and leaves $60^{\circ}$ at $B$ . By applying Pythagoras' theorem in right triangle $A B N$ , the length $A N$ is $\sqrt{3} units$ . We then read off the sines, cosines, and tangents directly from triangle $A B N$ without approximations. Use Figure 1 to see how the bisector forms these exact side lengths and angles.

An equilateral triangle of side $2 units$ split by perpendicular bisector $A N$ forms two $30^{\circ} – 60^{\circ} – 90^{\circ}$ right triangles.
By Pythagoras' theorem, the perpendicular bisector $A N = \sqrt{2^{2} - 1^{2}} = \sqrt{3} units$ in exact surd form.
From triangle $A B N$ , $\sin 30^{\circ} = \frac{1}{2}$ , $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ , and $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ .
For the $60^{\circ}$ angle, $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ , $\cos 60^{\circ} = \frac{1}{2}$ , and $\tan 60^{\circ} = \sqrt{3}$ .

Worked example. Base length

Y K = \frac{1}{2} \times 2 = 1 unit

, and hypotenuse

X Y = 2 units

. By Pythagoras' theorem,

X K^{2} + 1^{2} = 2^{2} ⟹ X K = \sqrt{4 - 1} = \sqrt{3} units

. In right-angled triangle

X Y K

, angle

Y X K = 30^{\circ}

, so

\tan 30^{\circ} = \frac{opposite}{adjacent} = \frac{1}{\sqrt{3}}

. Angle

X Y K = 60^{\circ}

, so

\tan 60^{\circ} = \frac{opposite}{adjacent} = \frac{\sqrt{3}}{1} = \sqrt{3}

. Answer: $X K = \sqrt{3} units$ , $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ , and $\tan 60^{\circ} = \sqrt{3}$ .

Check question: Using right-angled triangle $A B N$ where hypotenuse $A B = 2$ , base $B N = 1$ , and perpendicular $A N = \sqrt{3}$ , what is the exact value of $\cos 30^{\circ}$ given that $∠ B A N = 30^{\circ}$ ?

$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{1}{\sqrt{3}}$
$\frac{2}{\sqrt{3}}$

Show answer

$\frac{\sqrt{3}}{2}$ . In right-angled triangle $A B N$ , $\cos 30^{\circ} = \frac{adjacent}{hypotenuse} = \frac{A N}{A B} = \frac{\sqrt{3}}{2}$ .

Slide 2 · Summary and Evaluation of Ratios for 30° and 60°Model · 14 min

Figure 2: A summary table of trigonometric ratios for 30 and 60 degrees is shown alongside a step-by-step evaluation of a mathematical expression.

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Trigonometric Ratios of Special Angles: 30° and 60°

By the end of this lesson, learners should be able to

Key vocabulary

Lesson preview

Teach the full lesson on the projector

All 11 lessons in Trigonometry 1