Trigonometry

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Ratios and standard values

⏱ 3 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

In a right triangle, each acute angle has six ratios built from the three sides. Three are primary: sine, cosine, tangent; three are their reciprocals. Five standard angles cover almost every exam question, and one known ratio produces all six.

The six ratios

For an acute angle $\theta$ in a right triangle:

$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$
$\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}$
$\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}$

The other three flip these: $\text{cosec}=\dfrac{1}{\sin}$ , $\sec=\dfrac{1}{\cos}$ , $\cot=\dfrac{1}{\tan}$ .

Rule: Name the sides from the angle you are using. The opposite side changes when the angle changes.

The standard table, built not memorised

Write $\sin$ as $\dfrac{\sqrt{k}}{2}$ for $k=0,1,2,3,4$ :

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin$	$0$	$\dfrac12$	$\dfrac{1}{\sqrt2}$	$\dfrac{\sqrt3}{2}$	$1$
$\cos$	$1$	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{\sqrt2}$	$\dfrac12$	$0$

$\cos$ is $\sin$ read backwards. Then $\tan=\dfrac{\sin}{\cos}$ gives $0,\dfrac{1}{\sqrt3},1,\sqrt3$ and undefined at $90^\circ$ .

Read $\tan$ straight off that division: it starts at $0$ , passes $1$ exactly at $45^\circ$ , and blows up where cosine hits zero. A one-line check: $\tan30^\circ\tan60^\circ=1$ , which the table confirms because the angles are complementary.

Tip: Recompute the row from $\sqrt{k}/2$ instead of memorising. It cannot be forgotten in the hall.

One ratio makes all six

Given $\sin\theta=\dfrac35$ : opposite $3$ , hypotenuse $5$ , so adjacent $4$ (the $3$ - $4$ - $5$ triangle). Then $\cos=\dfrac45$ , $\tan=\dfrac34$ , $\text{cosec}=\dfrac53$ , $\sec=\dfrac54$ , $\cot=\dfrac43$ .

The Pythagoras step is the whole work: two sides known means the third follows.

Try the same move with $\cos\theta=\dfrac{5}{13}$ : adjacent $5$ , hypotenuse $13$ , so the opposite side is $12$ from the $5$ - $12$ - $13$ triplet. Then $\tan\theta=\dfrac{12}{5}$ , $\sec\theta=\dfrac{13}{5}$ , $\cot\theta=\dfrac{5}{12}$ and $\cosec\theta=\dfrac{13}{12}$ . Six answers come from one fact.

Watch: Keep the given ratio's fraction exact. Decimal sides break the triplets ( $3$ - $4$ - $5$ , $5$ - $12$ - $13$ , $20$ - $21$ - $29$ ).

Labelling a given triangle

Triangle right-angled at $B$ with $AB=5$ , $BC=12$ : hypotenuse $AC=13$ . For angle $A$ , the opposite side is $BC=12$ , so $\sin A=\dfrac{12}{13}$ .

Switch to angle $C$ and the roles swap: $\sin C=\dfrac{5}{13}$ . One drawing, two readings.

Reading an angle off a value

$2\sin\theta=\sqrt3$ means $\sin\theta=\dfrac{\sqrt3}{2}$ , which is the $60^\circ$ entry. Equations of this shape just run the table backwards.

The reverse lookup needs the function named first: the value $\dfrac12$ means $30^\circ$ for sine but $60^\circ$ for cosine, while $\dfrac{1}{\sqrt2}$ means $45^\circ$ for both. Options are often rationalised, so know $\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}$ too.

Remember: Every ratio value in this range belongs to exactly one standard angle, so match it and stop.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Standard values at 0/30/45/60/90

How to spot it:

A direct expression in standard angles.

Method

Replace each ratio by its table value.
Simplify the fractions.
Watch $45^\circ$ entries: $\tan45^\circ=1$ .

Why it works:

The five standard values turn any such expression into arithmetic within a line or two.

Try this

The value of $\sin 30^\circ + \cos 60^\circ$ is:

Show solution

$\sin30^\circ=\dfrac12$ , $\cos60^\circ=\dfrac12$ .
$\dfrac12+\dfrac12=1$ .

Answer

Type 2very common3 practice Q

One given ratio to the other ratios

How to spot it:

A single ratio given; another asked.

Method

Put the two known sides into a right triangle.
Find the third side by Pythagoras.
Read the asked ratio off the triangle.

Why it works:

All six ratios live on one triangle, so one ratio plus Pythagoras unlocks the rest.

Try this

If $\sin\theta = \frac{3}{5}$ ( $\theta$ acute), the value of $\tan\theta$ is:

Show solution

Adjacent $=\sqrt{25-9}=4$ .
$\tan\theta=\dfrac{3}{4}$ .

Answer

3/4

Type 3common2 practice Q

Two sides of the triangle to ratios

How to spot it:

A labelled right triangle with two side lengths.

Method

Find the third side if needed.
Identify the side roles for the asked angle.
Write the ratio.

Why it works:

Naming sides per angle converts geometry into a plain fraction.

Try this

In a triangle right-angled at B, AB = 5 cm and BC = 12 cm. The value of sin A is:

Show solution

$AC=\sqrt{25+144}=13$ .
Opposite $A$ is $BC=12$ .
$\sin A=\dfrac{12}{13}$ .

Answer

12/13

Type 4common2 practice Q

Mixed-function combinations

How to spot it:

Tangent additions like the tan-of-sum shape.

Method

Insert standard values.
Simplify surds by rationalising.
Recognise the result as a known ratio if it matches one.

Why it works:

These collapse to a single table value once the arithmetic settles.

Try this

The value of $\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}$ is:

Show solution

Numerator: $\sqrt3-\dfrac{1}{\sqrt3}=\dfrac{2}{\sqrt3}$ ; denominator: $1+1=2$ .
$\dfrac{2/\sqrt3}{2}=\dfrac{1}{\sqrt3}=\tan30^\circ$ .

Answer

1/sqrt(3)

Type 5common

Finding the angle from a ratio

How to spot it:

An equation with one unknown angle.

Method

Isolate the single ratio on one side.
Match the value to the table.
State the angle.

Why it works:

Within the standard range each value belongs to one angle, so solving is matching.

Try this

If $2\sin\theta=\sqrt3$ ( $0^\circ\le\theta\le90^\circ$ ), then $\theta$ is:

Show solution

$\sin\theta=\dfrac{\sqrt3}{2}$ .
That is the $60^\circ$ entry.

Answer

60 degrees

Formula sheet

Primary ratios

\sin\theta=\frac{o}{h},\quad \cos\theta=\frac{a}{h},\quad \tan\theta=\frac{o}{a}

o = opposite, a = adjacent, h = hypotenuse.

Reciprocals

\text{cosec}=\frac{1}{\sin},\quad \sec=\frac{1}{\cos},\quad \cot=\frac{1}{\tan}

Flip the fraction.

Standard values

\sin\theta=\frac{\sqrt{k}}{2},\ k=0,1,2,3,4

For 0, 30, 45, 60, 90 degrees; cos runs the row backwards.

One ratio to all

\sin\theta=\frac{3}{5}\Rightarrow\cos=\frac45,\ \tan=\frac34

Draw the 3-4-5 triangle and read every ratio off it.

Shortcuts that save time

⚡ The root-k-over-2 row

sin at 0, 30, 45, 60, 90 is root-0, root-1, root-2, root-3, root-4, all over 2. Cos is the same row reversed.

Example

The value of $\sin 30^\circ + \cos 60^\circ$ is:

Show solution

$\sin30^\circ=\dfrac12$ and $\cos60^\circ=\dfrac12$ .
$\dfrac12+\dfrac12=1$ .

Answer

⚡ Triplet finishes the ratios

Given sin = 3/5, place 3 and 5 in the triangle; the third side 4 completes 3-4-5 and every ratio follows.

Example

If $\sin\theta = \frac{3}{5}$ ( $\theta$ acute), the value of $\tan\theta$ is:

Show solution

Adjacent $=\sqrt{25-9}=4$ .
$\tan\theta=\dfrac{3}{4}$ .

Answer

3/4

⚡ Angle from a value

Isolate the ratio, then match the table entry. 2 sin = root 3 means sin = root-3 over 2, the 60-degree slot.

Example

If $2\sin\theta=\sqrt3$ ( $0^\circ\le\theta\le90^\circ$ ), then $\theta$ is:

Show solution

$\sin\theta=\dfrac{\sqrt3}{2}$ .
That value sits at $60^\circ$ .

Answer

60 degrees

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Taking the adjacent side as the hypotenuse.

The hypotenuse faces the right angle and is the longest side. In 3-4-5, $\sin=\dfrac35$ uses 5 as hypotenuse.

Mistake 02

Swapping opposite and adjacent when the angle changes.

Sides are named per angle: $\sin A=\dfrac{12}{13}$ but $\sin C=\dfrac{5}{13}$ in the same triangle.

Mistake 03

Writing $\cos 60^\circ=\dfrac{\sqrt3}{2}$ .

Cosine reverses the row: $\cos60^\circ=\dfrac12$ ; $\dfrac{\sqrt3}{2}$ is $\cos30^\circ$ .

Mistake 04

Mixing $\sec$ and $\text{cosec}$ .

$\sec$ pairs with $\cos$ ; $\text{cosec}$ pairs with $\sin$ . Both flip, nothing more.

Mistake 05

Using decimals for the $3$ - $4$ - $5$ family.

Exact fractions keep the triplet visible and the arithmetic short.

Quick revision

Read this the night before the exam.

$\sin=\dfrac{opp}{hyp}$ , $\cos=\dfrac{adj}{hyp}$ , $\tan=\dfrac{opp}{adj}$ ; the rest are reciprocals.
$\sin$ row: $\dfrac{\sqrt{k}}{2}$ for $k=0..4$ ; $\cos$ is the row reversed.
$\tan45^\circ=1$ ; $\tan$ is undefined at $90^\circ$ .
One ratio plus Pythagoras yields all six; keep triplets exact.
Label opposite and adjacent afresh for each angle.
An equation like $2\sin\theta=\sqrt3$ is a table lookup: $\theta=60^\circ$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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