Trigonometric ratios of acute angles

Mr Tư leans a $5$ m ladder against a wall, with its foot $3$ m from the bottom of the wall. Safety advice says the angle between a ladder and the ground should be about $75^\circ$ . Is Mr Tư's ladder steep enough?

We know two sides of a right-angled triangle, but we need an angle. This lesson gives us the tool that links sides to angles: the trigonometric ratios.

The ladder against the wall

The trigonometric ratios of an acute angle

Explore: the ratio does not depend on the size of the triangle

Take two right-angled triangles with the same acute angle $\alpha$ at vertex $B$ : the first has sides $3$ , $4$ , $5$ ; the second has sides $6$ , $8$ , $10$ (twice as big). We work out the ratio of the side opposite $\alpha$ to the hypotenuse:

Triangle	opposite	hypotenuse	opposite : hypotenuse
first	$3$	$5$	$0.6$
second	$6$	$10$	$0.6$

The two triangles are similar (grade 8), so the ratios are the same. The ratio depends only on the angle $\alpha$ , not on how big the triangle is.

The opposite side, adjacent side and hypotenuse for angle

\alpha

Because the hypotenuse is the longest side, we always have $0<\sin\alpha<1$ and $0<\cos\alpha<1$ . A well-known memory aid is SOH CAH TOA (Sine = Opposite : Hypotenuse, Cosine = Adjacent : Hypotenuse, Tangent = Opposite : Adjacent); the cotangent is the tangent turned upside down.

Example Finding the four ratios

Triangle $ABC$ has a right angle at $A$ , $AB=8$ cm and $AC=6$ cm. Find the trigonometric ratios of angle $B$ .

Solution. In the right-angled triangle $ABC$ , $BC^2=AB^2+AC^2$ (Pythagoras' theorem). For angle $B$ , the opposite side is $AC$ and the adjacent side is $AB$ . Hence

$BC=\sqrt{8^2+6^2}=\sqrt{100}=10;\qquad \sin B=\frac{6}{10}=\frac35,\quad \cos B=\frac{8}{10}=\frac45,\quad \tan B=\frac68=\frac34,\quad \cot B=\frac86=\frac43.$

So $\sin B=\dfrac35$ , $\cos B=\dfrac45$ , $\tan B=\dfrac34$ , $\cot B=\dfrac43$ .

The opposite and adjacent sides change when we look at the other acute angle, so always identify them again before writing a ratio.

The ratios of special angles

Half of a square with side $1$ is a right-angled isosceles triangle with two $45^\circ$ angles and hypotenuse $\sqrt2$ . Half of an equilateral triangle with side $2$ is a right-angled triangle with angles $30^\circ$ and $60^\circ$ and sides $1$ , $\sqrt3$ , $2$ . These two triangles give the table below.

$\alpha$	$30^\circ$	$45^\circ$	$60^\circ$
$\sin\alpha$	$\dfrac12$	$\dfrac{\sqrt2}2$	$\dfrac{\sqrt3}2$
$\cos\alpha$	$\dfrac{\sqrt3}2$	$\dfrac{\sqrt2}2$	$\dfrac12$
$\tan\alpha$	$\dfrac{\sqrt3}3$	$1$	$\sqrt3$
$\cot\alpha$	$\sqrt3$	$1$	$\dfrac{\sqrt3}3$

For example, $\sin45^\circ=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$ (rationalising the denominator as in the previous chapter).

The ratios of complementary angles

In triangle $ABC$ with a right angle at $A$ , angles $B$ and $C$ are complementary: $\angle B+\angle C=90^\circ$ . The side opposite $B$ is the side adjacent to $C$ , so $\sin B=\dfrac{AC}{BC}=\cos C$ .

Example Using complementary angles

a) Write $\sin72^\circ$ and $\tan65^\circ$ as trigonometric ratios of angles smaller than $45^\circ$ .

b) Without a calculator, evaluate $\dfrac{\sin25^\circ}{\cos65^\circ}$ and $\tan40^\circ\cdot\tan50^\circ$ .

Solution. We have $72^\circ+18^\circ=90^\circ$ , $65^\circ+25^\circ=90^\circ$ , $25^\circ+65^\circ=90^\circ$ , $40^\circ+50^\circ=90^\circ$ (complementary angles) and $\tan40^\circ\cdot\cot40^\circ=\dfrac{\text{opp}}{\text{adj}}\cdot\dfrac{\text{adj}}{\text{opp}}=1$ . Hence

$\sin72^\circ=\cos18^\circ,\quad \tan65^\circ=\cot25^\circ;\qquad \frac{\sin25^\circ}{\cos65^\circ}=\frac{\sin25^\circ}{\sin25^\circ}=1,\quad \tan40^\circ\cdot\tan50^\circ=\tan40^\circ\cdot\cot40^\circ=1.$

So a) $\cos18^\circ$ and $\cot25^\circ$ ; b) both expressions equal $1$ .

For any other angle, we use a calculator (set to degree mode, shown as D). The sin, cos and tan keys give the ratio; SHIFT + sin (or cos, tan) gives the angle from a ratio. There is no cot key, so we use $\cot\alpha=\dfrac{1}{\tan\alpha}$ .

Watch out Common mistakes

Mixing up opposite and adjacent. Wrong: in a triangle with a right angle at $A$ , $\sin C=\dfrac{AC}{BC}$ . Why it is wrong: the side opposite $C$ is $AB$ . Right: $\sin C=\dfrac{AB}{BC}$ .
Putting the hypotenuse on top. Wrong: $\cos B=\dfrac{BC}{AB}$ . Why it is wrong: this ratio is greater than $1$ , but $\cos B<1$ . Right: $\cos B=\dfrac{AB}{BC}$ .
Mixing up special values. Wrong: $\sin60^\circ=\dfrac12$ . Why it is wrong: that is $\sin30^\circ$ (the side opposite $30^\circ$ is the shortest). Right: $\sin60^\circ=\dfrac{\sqrt3}2$ .
Using the wrong calculator mode. Wrong: $\sin30$ gives $-0.988$ . Why it is wrong: the calculator is in radian mode. Right: switch to degree mode to get $0.5$ .

Key terms

English	Tiếng Việt
trigonometric ratio	tỉ số lượng giác
acute angle	góc nhọn
hypotenuse	cạnh huyền
opposite side	cạnh đối
adjacent side	cạnh kề
sine, cosine, tangent, cotangent	sin, côsin, tang, côtang
complementary angles	hai góc phụ nhau
right-angled isosceles triangle	tam giác vuông cân

Summary

$\sin\alpha=\dfrac{\text{opp}}{\text{hyp}}$ , $\cos\alpha=\dfrac{\text{adj}}{\text{hyp}}$ , $\tan\alpha=\dfrac{\text{opp}}{\text{adj}}$ , $\cot\alpha=\dfrac{\text{adj}}{\text{opp}}$ .
The ratios depend only on the angle; $0<\sin\alpha<1$ , $0<\cos\alpha<1$ ; $\tan\alpha\cdot\cot\alpha=1$ .
$\sin30^\circ=\cos60^\circ=\dfrac12$ , $\sin45^\circ=\cos45^\circ=\dfrac{\sqrt2}2$ , $\sin60^\circ=\cos30^\circ=\dfrac{\sqrt3}2$ , $\tan45^\circ=1$ , $\tan60^\circ=\sqrt3$ , $\tan30^\circ=\dfrac{\sqrt3}3$ .
Complementary angles: the sine of one is the cosine of the other, and the tangent of one is the cotangent of the other.
Calculator: degree mode; SHIFT + sin/cos/tan to find an angle.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Triangle $ABC$ has a right angle at $A$ , $AB=3$ and $AC=4$ . Find the trigonometric ratios of angle $B$ .

What you need and how to start

What you need: the definitions of the four ratios; Pythagoras' theorem.

Steps:

Find $BC$ .
For angle $B$ : the opposite side is $AC$ , the adjacent side is $AB$ .

Show solution

In the right-angled triangle $ABC$ , $BC=\sqrt{AB^2+AC^2}=\sqrt{9+16}=5$ (Pythagoras' theorem). Hence $\sin B=\dfrac{AC}{BC}=\dfrac45$ , $\cos B=\dfrac{AB}{BC}=\dfrac35$ , $\tan B=\dfrac{AC}{AB}=\dfrac43$ , $\cot B=\dfrac{AB}{AC}=\dfrac34$ .

Answer. $\dfrac45$ ; $\dfrac35$ ; $\dfrac43$ ; $\dfrac34$ .

Exercise 2

Triangle $ABC$ has a right angle at $A$ , $BC=17$ and $AB=8$ . Find the trigonometric ratios of angle $C$ .

What you need and how to start

What you need: $AC^2=BC^2-AB^2$ ; for angle $C$ , the opposite side is $AB$ .

Steps:

Find $AC$ .
Write the four ratios.

Show solution

In the right-angled triangle $ABC$ , $AC=\sqrt{BC^2-AB^2}=\sqrt{289-64}=15$ (Pythagoras' theorem). Hence $\sin C=\dfrac{AB}{BC}=\dfrac8{17}$ , $\cos C=\dfrac{AC}{BC}=\dfrac{15}{17}$ , $\tan C=\dfrac{AB}{AC}=\dfrac8{15}$ , $\cot C=\dfrac{AC}{AB}=\dfrac{15}8$ .

Answer. $\dfrac8{17}$ ; $\dfrac{15}{17}$ ; $\dfrac8{15}$ ; $\dfrac{15}8$ .

Exercise 3

Evaluate: a) $\sin45^\circ\cdot\cos45^\circ$ ; b) $\tan60^\circ-\cot30^\circ+\tan45^\circ$ .

What you need and how to start

What you need: the table of special-angle ratios.

Steps:

Substitute the values.
Work it out.

Show solution

a) $\sin45^\circ\cdot\cos45^\circ=\dfrac{\sqrt2}2\cdot\dfrac{\sqrt2}2=\dfrac24=\dfrac12$ . b) $\tan60^\circ-\cot30^\circ+\tan45^\circ=\sqrt3-\sqrt3+1=1$ .

Answer. a) $\dfrac12$ ; b) $1$ .

Exercise 4

Write as trigonometric ratios of angles smaller than $45^\circ$ : $\sin70^\circ$ , $\cos58^\circ$ , $\tan80^\circ$ , $\cot47^\circ$ .

What you need and how to start

What you need: if $\alpha+\beta=90^\circ$ , then $\sin\alpha=\cos\beta$ and $\tan\alpha=\cot\beta$ .

Steps:

Find the complement of each angle.
Swap sin ↔ cos and tan ↔ cot.

Show solution

Since the pairs $70^\circ$ and $20^\circ$ , $58^\circ$ and $32^\circ$ , $80^\circ$ and $10^\circ$ , $47^\circ$ and $43^\circ$ are complementary, $\sin70^\circ=\cos20^\circ$ , $\cos58^\circ=\sin32^\circ$ , $\tan80^\circ=\cot10^\circ$ , $\cot47^\circ=\tan43^\circ$ .

Answer. $\cos20^\circ$ ; $\sin32^\circ$ ; $\cot10^\circ$ ; $\tan43^\circ$ .

Exercise 5

Without a calculator, evaluate $M=\dfrac{\sin32^\circ}{\cos58^\circ}+\tan17^\circ\cdot\tan73^\circ$ .

What you need and how to start

What you need: complementary angles and $\tan\alpha\cdot\cot\alpha=1$ .

Steps:

Write $\cos58^\circ=\sin32^\circ$ .
Write $\tan73^\circ=\cot17^\circ$ .

Show solution

Since $58^\circ+32^\circ=90^\circ$ and $73^\circ+17^\circ=90^\circ$ , $\cos58^\circ=\sin32^\circ$ and $\tan73^\circ=\cot17^\circ$ . Hence $M=\dfrac{\sin32^\circ}{\sin32^\circ}+\tan17^\circ\cdot\cot17^\circ=1+1=2$ .

Answer. $M=2$ .

Exercise 6

Use a calculator:

a) Find $\sin54^\circ$ and $\tan20^\circ$ (rounded to four decimal places).

b) Find the acute angle $\alpha$ (to the nearest degree) given that $\tan\alpha=2$ ; given that $\sin\alpha=0.25$ .

What you need and how to start

What you need: degree mode; SHIFT + tan (or sin) to find an angle.

Steps:

Check that the calculator is in mode D.
Key in and round.

Show solution

a) Using a calculator, we get $\sin54^\circ\approx0.8090$ and $\tan20^\circ\approx0.3640$ .

b) From $\tan\alpha=2$ we get $\alpha\approx63^\circ$ . From $\sin\alpha=0.25$ we get $\alpha\approx14^\circ$ .

Answer. a) $0.8090$ ; $0.3640$ ; b) $63^\circ$ ; $14^\circ$ .

Exercise 7

A " $10\%$ gradient" road sign means that for every $100$ m horizontally, the road rises $10$ m. Building guidance says a wheelchair ramp should be no steeper than $1:12$ (a rise of $1$ m over $12$ m horizontally). Find the angle each slope makes with the horizontal (to the nearest degree).

What you need and how to start

What you need: the angle of slope $\alpha$ has $\tan\alpha=\dfrac{\text{rise}}{\text{horizontal distance}}$ .

Steps:

Write $\tan\alpha$ for each slope.
Press SHIFT tan.

Show solution

Let $\alpha$ and $\beta$ be the angles of the $10\%$ road and of the wheelchair ramp. We have $\tan\alpha=\dfrac{10}{100}=0.1$ , so $\alpha\approx6^\circ$ ; and $\tan\beta=\dfrac1{12}$ , so $\beta\approx5^\circ$ .

Answer. About $6^\circ$ and $5^\circ$ .

Exercise 8

Triangle $ABC$ has a right angle at $A$ , $BC=10$ cm and $\sin B=0.6$ .

a) Find $AC$ .

b) Find $AB$ .

c) Find $\tan B$ and $\cos C$ .

d) Find the size of angle $B$ (to the nearest degree).

What you need and how to start

What you need: $\sin B=\dfrac{AC}{BC}$ ; Pythagoras' theorem; $\cos C=\sin B$ because the angles are complementary.

Steps:

$AC=BC\cdot\sin B$ .
$AB=\sqrt{BC^2-AC^2}$ .
Write the ratios; press SHIFT sin $0.6$ .

Show solution

a) In the right-angled triangle $ABC$ , $\sin B=\dfrac{AC}{BC}$ , so $AC=BC\cdot\sin B=10\cdot0.6=6$ (cm).

b) $AB=\sqrt{BC^2-AC^2}=\sqrt{100-36}=8$ (cm) (Pythagoras' theorem).

c) $\tan B=\dfrac{AC}{AB}=\dfrac68=0.75$ ; $\cos C=\sin B=0.6$ (complementary angles).

d) From $\sin B=0.6$ we get $\angle B\approx37^\circ$ .

Answer. a) $6$ cm; b) $8$ cm; c) $0.75$ and $0.6$ ; d) $37^\circ$ .

Exercise 9

Triangle $ABC$ has a right angle at $A$ , $AB=1$ and $AC=2$ . Find $\sin B$ and $\cos B$ , with denominators free of roots.

What you need and how to start

What you need: rationalising the denominator (see transforming and simplifying expressions with square roots).

Steps:

Find $BC=\sqrt5$ .
Write the ratios, then rationalise the denominators.

Show solution

In the right-angled triangle $ABC$ , $BC=\sqrt{AB^2+AC^2}=\sqrt{1+4}=\sqrt5$ (Pythagoras' theorem). Hence $\sin B=\dfrac{AC}{BC}=\dfrac2{\sqrt5}=\dfrac{2\sqrt5}5$ and $\cos B=\dfrac{AB}{BC}=\dfrac1{\sqrt5}=\dfrac{\sqrt5}5$ .

Answer. $\sin B=\dfrac{2\sqrt5}5$ , $\cos B=\dfrac{\sqrt5}5$ .

Exercise 10 Challenge

Triangle $ABC$ is isosceles with $AB=AC$ , $BC=12$ cm and $\angle B=30^\circ$ . Draw the altitude $AH$ . Find $AH$ and $AB$ (in a neat exact form).

What you need and how to start

What you need: in an isosceles triangle the altitude is also a median, so $BH=6$ cm; $\tan30^\circ=\dfrac{\sqrt3}3$ , $\cos30^\circ=\dfrac{\sqrt3}2$ .

Steps:

In the right-angled triangle $ABH$ : $\tan B=\dfrac{AH}{BH}$ .
$\cos B=\dfrac{BH}{AB}$ gives $AB$ ; rationalise the denominator.

Show solution

Since triangle $ABC$ is isosceles with $AB=AC$ , the altitude $AH$ is also a median, hence $BH=\dfrac{BC}2=6$ cm.

In triangle $ABH$ with a right angle at $H$ , $AH=BH\cdot\tan B=6\cdot\dfrac{\sqrt3}3=2\sqrt3$ (cm) and $AB=\dfrac{BH}{\cos B}=\dfrac{6\cdot2}{\sqrt3}=\dfrac{12}{\sqrt3}=4\sqrt3$ (cm).

Answer. $AH=2\sqrt3$ cm, $AB=4\sqrt3$ cm.

a) $BC=2$ .	True	False
b) $\sin B=\dfrac12$ .	True	False
c) $\tan B=\sqrt3$ .	True	False
d) $\angle C=60^\circ$ .	True	False

The trigonometric ratios of an acute angle

Explore: the ratio does not depend on the size of the triangle

The ratios of special angles

The ratios of complementary angles

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer