AFCAT Numerical Ability · Heights and Distances

The Trigonometric Ratios at 0°, 30°, 45°, 60° and 90°

Sine, cosine and tangent are ratios of the sides of a right triangle; at the five standard angles their values come from two set squares and are learnt by heart.

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Why this matters

More than a third of the chapter's questions sit on this page, and most of them ask one value, or a sum or product of two. The HARD ones give one ratio as a fraction and ask for another. Marks are lost by putting √3/2 at 30° instead of 60°, or by pairing cosec with cos.

Concept 1 of 5: Sine, cosine and tangent at the standard angles

In a right triangle, take one acute angle

\theta

. The side across from it is the opposite side, the side beside it is the adjacent side, and the longest side is the hypotenuse. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. The values at 30°, 45° and 60° come from two set squares: the half square with sides

1, 1, \sqrt2

and the half equilateral triangle with sides

1, \sqrt3, 2

Definition

Sine $= \dfrac{\text{opposite}}{\text{hypotenuse}}$ , cosine $= \dfrac{\text{adjacent}}{\text{hypotenuse}}$ , tangent $= \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sin\theta}{\cos\theta}$ .
A memory line for sine at 0°, 30°, 45°, 60°, 90°: $\dfrac{\sqrt0}{2}, \dfrac{\sqrt1}{2}, \dfrac{\sqrt2}{2}, \dfrac{\sqrt3}{2}, \dfrac{\sqrt4}{2}$ . Cosine is the same list read backwards.
$\sin^2\theta$ means $(\sin\theta)^2$ : square the value, not the angle.
Sine and tangent grow from 0° to 90°; cosine falls.

Angle	sin	cos	tan
0°	$0$	$1$	$0$
30°	$\tfrac12$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt3}$
45°	$\tfrac{1}{\sqrt2}$	$\tfrac{1}{\sqrt2}$	$1$
60°	$\tfrac{\sqrt3}{2}$	$\tfrac12$	$\sqrt3$
90°	$1$	$0$	not defined tan 90° would be 1 divided by 0, which has no value.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Heights and Distances · Trigonometric Ratios

What is the value of

\sin(60°) \times \cos(30°)

√3/2 belongs to 60° for sine

\sin 60^\circ = \tfrac{\sqrt3}{2}

and

\sin 30^\circ = \tfrac12

. Swapping them is the most common slip; the memory line puts the bigger value at the bigger angle.

Square the value, not the angle

\sin^2 45^\circ = \left(\tfrac{1}{\sqrt2}\right)^2 = \tfrac12

. It is not

\sin 45^\circ

doubled, and it is not the sine of some larger angle.

Drill 10 more on sine, cosine and tangent at the standard angles

Concept 2 of 5: Cosecant, secant and cotangent

Each of the other three ratios is one of the first three turned upside down. Cosecant is 1 over sine, secant is 1 over cosine, and cotangent is 1 over tangent. Flip the value from the first table and simplify.

Definition

$\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$ , $\sec\theta = \dfrac{1}{\cos\theta}$ , $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$ .
A ratio times its reciprocal is 1: $\sin\theta \times \operatorname{cosec}\theta = 1$ .
If one is given as a fraction, the other is the same fraction flipped.
Options often print $\tfrac{2}{\sqrt3}$ as $\tfrac{2\sqrt3}{3}$ : multiply top and bottom by $\sqrt3$ to match.

Angle	cosec	sec	cot
0°	not defined	$1$	not defined
30°	$2$	$\tfrac{2}{\sqrt3}$	$\sqrt3$
45°	$\sqrt2$	$\sqrt2$	$1$
60°	$\tfrac{2}{\sqrt3}$	$2$	$\tfrac{1}{\sqrt3}$
90°	$1$	not defined	$0$

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Heights and Distances · Trigonometric Ratios

\cosec(\theta) = \frac{13}{5}

, then what is the value of

\sin(\theta)

Cosec goes with sine, sec with cosine

The names mislead:

\operatorname{cosec}\theta = \tfrac{1}{\sin\theta}

, not

\tfrac{1}{\cos\theta}

. So

\operatorname{cosec} 30^\circ = 2

and

\sec 30^\circ = \tfrac{2}{\sqrt3}

Reciprocal is not negative

The reciprocal of

\tfrac{7}{9}

\tfrac{9}{7}

. Writing

-\tfrac{7}{9}

1 - \tfrac{7}{9}

confuses flipping with other operations.

Drill 5 more on cosecant, secant and cotangent

Concept 3 of 5: Finding the angle from a value

Read the two tables backwards. If a ratio of an acute angle has a standard value, the angle is the one that value sits under. Once the angle is known, any other ratio of it is one more look at the table.

Definition

First make the ratio stand alone: $2\sin\theta = 1$ means $\sin\theta = \tfrac12$ .
A reciprocal ratio is flipped first: $\operatorname{cosec}\theta = \sqrt2$ means $\sin\theta = \tfrac{1}{\sqrt2}$ .
If two ratios are equal, divide: $\sin\theta = \cos\theta$ gives $\tan\theta = 1$ .
In this chapter the angle is acute (between 0° and 90°), so each value names exactly one angle.

Reading the table backwards

\sin\theta = \tfrac{1}{\sqrt2} \;\Rightarrow\; \theta = 45^\circ

Worked example

\operatorname{cosec}\theta = \sqrt2

and

\theta

is acute, find

\theta

and

\cot\theta

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Heights and Distances · Trigonometric Ratios

\sec(\theta) = 2

, then what is the value of

\theta

Strip the number in front first

2\sin\theta = 1

means

\sin\theta = \tfrac12

and

\theta = 30^\circ

. Reading it as

\sin\theta = 2

gives no angle at all, since sine never passes 1.

Flip a reciprocal before looking it up

\cot\theta = \sqrt3

means

\tan\theta = \tfrac{1}{\sqrt3}

, so

\theta = 30^\circ

, not 60°. Looking

\sqrt3

up in the tangent column gives the complement.

Drill 5 more on finding the angle from a value

Concept 4 of 5: Complementary angles and the square identity

The two acute angles of a right triangle add up to 90°. The side opposite one is adjacent to the other, so the sine of one angle is the cosine of the other. And because the two shorter sides and the hypotenuse obey Pythagoras,

\sin^2\theta + \cos^2\theta

is always 1.

Definition

Complementary pairs: $\sin(90^\circ - \theta) = \cos\theta$ , $\cos(90^\circ - \theta) = \sin\theta$ , $\tan(90^\circ - \theta) = \cot\theta$ .
So $\tan\theta \times \tan(90^\circ - \theta) = \tan\theta \times \cot\theta = 1$ .
Square identity: $\sin^2\theta + \cos^2\theta = 1$ for every angle. Dividing by $\cos^2\theta$ gives $1 + \tan^2\theta = \sec^2\theta$ .
These answer a question without knowing $\theta$ : $\tfrac{\sin 40^\circ}{\cos 50^\circ} = 1$ .

Complement and square identity

\sin(90^\circ - \theta) = \cos\theta, \qquad \sin^2\theta + \cos^2\theta = 1

Worked example

Find

\dfrac{\tan 20^\circ}{\cot 70^\circ} + \sin^2 35^\circ + \cos^2 35^\circ

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Heights and Distances · Trigonometric Ratios

\cos(\theta) = \frac{1}{2}

, then what is the value of

\sin(90° - \theta)

Complementary means adding to 90°

\sin 20^\circ

equals

\cos 70^\circ

, not

\cos 20^\circ

. Check that the two angles add to 90° before swapping sine for cosine.

The identity needs the same angle

\sin^2 35^\circ + \cos^2 35^\circ = 1

, but

\sin^2 35^\circ + \cos^2 55^\circ

is not 1 (it is

2\sin^2 35^\circ

Drill 4 more on complementary angles and the square identity

Concept 5 of 5: All the ratios from one given ratio

A ratio given as a fraction names two sides of a right triangle. Draw it, find the third side by Pythagoras, and read off any other ratio. When both sine and cosine are given, tangent is simply one divided by the other.

Definition

$\tan\theta = \tfrac{8}{15}$ means opposite $8k$ , adjacent $15k$ ; then the hypotenuse is $\sqrt{8^2 + 15^2}\,k = 17k$ . The $k$ cancels in every ratio.
Triples worth knowing: $3, 4, 5$ ; $8, 15, 17$ ; $7, 24, 25$ ; and any multiple of them.
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ need no triangle.
For an acute angle every ratio is positive.

The third side

\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2

Worked example

\tan\theta = \tfrac{8}{15}

, find

\sin\theta

and

\cos\theta

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 5 · Heights and Distances · Trigonometric Ratios

What is the value of

\tan(\theta)

\sin(\theta) = \frac{5}{13}

and

\cos(\theta) = \frac{12}{13}

Put the numbers on the right sides

For

\cos\theta = \tfrac{15}{17}

the 15 is the adjacent side and 17 the hypotenuse. Putting 15 opposite the angle swaps sine and cosine in every answer.

The hypotenuse is the longest side

If the side you find comes out longer than the hypotenuse, you added squares where you should have subtracted: the missing shorter side is

\sqrt{\text{hyp}^2 - \text{side}^2}

Drill 3 more on all the ratios from one given ratio

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Finding the angle from a value
Reading the table backwards
$\sin\theta = \tfrac{1}{\sqrt2} \;\Rightarrow\; \theta = 45^\circ$
Complementary angles and the square identity
Complement and square identity
$\sin(90^\circ - \theta) = \cos\theta, \qquad \sin^2\theta + \cos^2\theta = 1$
All the ratios from one given ratio
The third side
$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

Reference tables (2)

Sine, cosine and tangent at the standard angles5 rows

Angle	sin	cos	tan
0°	$0$	$1$	$0$
30°	$\tfrac12$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt3}$
45°	$\tfrac{1}{\sqrt2}$	$\tfrac{1}{\sqrt2}$	$1$
60°	$\tfrac{\sqrt3}{2}$	$\tfrac12$	$\sqrt3$
90°	$1$	$0$	not defined tan 90° would be 1 divided by 0, which has no value.

Cosecant, secant and cotangent5 rows

Angle	cosec	sec	cot
0°	not defined	$1$	not defined
30°	$2$	$\tfrac{2}{\sqrt3}$	$\sqrt3$
45°	$\sqrt2$	$\sqrt2$	$1$
60°	$\tfrac{2}{\sqrt3}$	$2$	$\tfrac{1}{\sqrt3}$
90°	$1$	not defined	$0$

Watch out for (10)

√3/2 belongs to 60° for sine
→ Sine, cosine and tangent at the standard angles
Square the value, not the angle
→ Sine, cosine and tangent at the standard angles
Cosec goes with sine, sec with cosine
→ Cosecant, secant and cotangent
Reciprocal is not negative
→ Cosecant, secant and cotangent
Strip the number in front first
→ Finding the angle from a value
Flip a reciprocal before looking it up
→ Finding the angle from a value
Complementary means adding to 90°
→ Complementary angles and the square identity
The identity needs the same angle
→ Complementary angles and the square identity
Put the numbers on the right sides
→ All the ratios from one given ratio
The hypotenuse is the longest side
→ All the ratios from one given ratio

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