Sec 2 · G3 (Express)

Trigonometric ratios

Practise trigonometric ratios with step-by-step help. Free account.

1. Sine, cosine and tangent

A right-angled triangle with angle θ: the hypotenuse is opposite the right angle, the opposite side faces θ, and the adjacent side is next to θ.

Ratio	Formula
sine	$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
cosine	$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
tangent	$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$

2. Finding a side

Hypotenuse 10 cm, angle 35°: the side opposite

\displaystyle \sin 35^\circ = \dfrac{x}{10}

e.g.

\displaystyle \sin 35^\circ = \dfrac{x}{10}

\displaystyle x = 10\sin 35^\circ

\displaystyle x = 5.74\text{ (3 s.f.)}

The unknown on the bottom

\displaystyle \cos 50^\circ = \dfrac{8}{x}

e.g.

\displaystyle \cos 50^\circ = \dfrac{8}{x}

\displaystyle x = \dfrac{8}{\cos 50^\circ}

\displaystyle x = 12.4\text{ (3 s.f.)}

📖 Worked example

In the diagram, $\angle KLM = 90^\circ$ . Find $x$ , correct to 3 significant figures.

Step 1 of 3

Label the sides from the 67° angle

From the 67° angle: the side

x

cm is the opposite and the side

8.5

cm is the hypotenuse.

📖 Worked example

From a point $A$ on level ground $15$ m from the foot of an HDB block, the angle of elevation of the top $T$ is $49^\circ$ . Find the height of the HDB block, $h$ m, correct to 3 significant figures.

Step 1 of 3

Label the sides from the angle of elevation

From the angle of elevation: the height

h

m is the opposite and the distance

15

m along the ground is the adjacent.

📖 Worked example

In the diagram, $\angle WXY = 90^\circ$ and $\angle WYZ = 90^\circ$ . $WX = 6$ cm, $\angle XWY = 28^\circ$ , $\angle YWZ = 44^\circ$ and $YZ = x$ cm. Find $x$ , correct to 3 significant figures.

Step 1 of 4

Triangle WXY: label the sides from the 28° angle at W

In triangle

WXY

, from the 28° angle at W:

WY

is the hypotenuse and

WX = 6

cm is the adjacent.

📖 Worked example

An aircraft takes off and climbs in a straight line at $8^\circ$ to the horizontal, at a constant speed of $216$ km/h. Find the height it gains in $40$ seconds, $h$ m, correct to 3 significant figures.

Step 1 of 2

Change the speed to metres per second (÷ 3.6), then distance = speed × time

Speed

= \dfrac{216}{3.6}

= 60

m/sDistance

= 60 \times 40

= 2400

📖 Worked example

From a point $P$ on level ground $12$ m from the foot of a school hall, the angles of elevation of the top and the bottom of a window in the wall are $35^\circ$ and $27^\circ$ . Find the height of the window, $h$ m, correct to 3 significant figures.

Step 1 of 2

Each height is opposite its angle of elevation and the distance to the wall is adjacent: use tan for the bottom (B m) and the top (T m)

Bottom:

\tan 27^\circ = \dfrac{B}{12}

, so

B = 12 \tan 27^\circ

B = 6.1143\ldots

mTop:

\tan 35^\circ = \dfrac{T}{12}

, so

T = 12 \tan 35^\circ

T = 8.4024\ldots

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3. Finding an angle

Opposite 3, adjacent 5: use the inverse

\displaystyle \tan\theta = \dfrac{3}{5}

e.g.

\displaystyle \theta = \tan^{-1} \dfrac{3}{5}

\displaystyle \theta = 31.0^\circ\text{ (1 d.p.)}

📖 Worked example

In the diagram, $\angle DEF = 90^\circ$ . Find $x$ , correct to 1 decimal place.

Step 1 of 3

Label the two known sides from the angle x°

From the angle x°: the side

8

cm is the opposite and the side

12.5

cm is the hypotenuse.

📖 Worked example

In the diagram, $\angle STU = 90^\circ$ , $SU = 15$ cm and $TU = 9$ cm. Write down the value of $\sin \angle UST$ , giving your answer as a fraction in its simplest form.

Step 1 of 2

Label the two sides that sin uses, from ∠UST

From

\angle UST

TU

is the opposite and

SU

is the hypotenuse.

📖 Worked example

In the diagram, $\angle KLM = 90^\circ$ , $KM = 20$ cm and $\sin \angle LKM = \dfrac{4}{5}$ . The point $N$ lies on $LM$ such that $LN : NM = 1 : 3$ . Find the length of $LN$ .

Step 1 of 2

The ratio tells you the sides in proportion: sin uses LM and KM

\dfrac{LM}{KM} = \dfrac{4}{5}

(from the sin ratio)

LM = \dfrac{4}{5} \times 20

= 16

📖 Worked example

A ladder $4$ m long leans against an upright wall. Its foot is on level ground, $1.2$ m from the wall. The ladder is safe to use when the angle $x^\circ$ it makes with the ground is between $72^\circ$ and $78^\circ$ . Find $x$ and decide whether the ladder is safe to use.

Step 1 of 4

Label the two known sides from the angle x° at the foot of the ladder

From the angle x° at the foot of the ladder: the distance from the wall,

1.2

m, is the adjacent and the ladder,

4

m, is the hypotenuse.

Practise this →

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Practice questions by skill

1. Sine, cosine and tangent

Ratio	Formula
sine	$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
cosine	$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
tangent	$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$

2. Finding a side

Hypotenuse 10 cm, angle 35°: the side opposite

\displaystyle \sin 35^\circ = \dfrac{x}{10}

e.g.

\displaystyle \sin 35^\circ = \dfrac{x}{10}

\displaystyle x = 10\sin 35^\circ

\displaystyle x = 5.74\text{ (3 s.f.)}

The unknown on the bottom

\displaystyle \cos 50^\circ = \dfrac{8}{x}

e.g.

\displaystyle \cos 50^\circ = \dfrac{8}{x}

\displaystyle x = \dfrac{8}{\cos 50^\circ}

\displaystyle x = 12.4\text{ (3 s.f.)}

📖 Worked example

In the diagram, $\angle KLM = 90^\circ$ . Find $x$ , correct to 3 significant figures.

Step 1 of 3

Label the sides from the 67° angle

From the 67° angle: the side

x

cm is the opposite and the side

8.5

cm is the hypotenuse.

📖 Worked example

From a point $A$ on level ground $15$ m from the foot of an HDB block, the angle of elevation of the top $T$ is $49^\circ$ . Find the height of the HDB block, $h$ m, correct to 3 significant figures.

Step 1 of 3

Label the sides from the angle of elevation

From the angle of elevation: the height

h

m is the opposite and the distance

15

m along the ground is the adjacent.

📖 Worked example

In the diagram, $\angle WXY = 90^\circ$ and $\angle WYZ = 90^\circ$ . $WX = 6$ cm, $\angle XWY = 28^\circ$ , $\angle YWZ = 44^\circ$ and $YZ = x$ cm. Find $x$ , correct to 3 significant figures.

Step 1 of 4

Triangle WXY: label the sides from the 28° angle at W

In triangle

WXY

, from the 28° angle at W:

WY

is the hypotenuse and

WX = 6

cm is the adjacent.

📖 Worked example

Step 1 of 2

Change the speed to metres per second (÷ 3.6), then distance = speed × time

Speed

= \dfrac{216}{3.6}

= 60

m/sDistance

= 60 \times 40

= 2400

📖 Worked example

Step 1 of 2

Each height is opposite its angle of elevation and the distance to the wall is adjacent: use tan for the bottom (B m) and the top (T m)

Bottom:

\tan 27^\circ = \dfrac{B}{12}

, so

B = 12 \tan 27^\circ

B = 6.1143\ldots

mTop:

\tan 35^\circ = \dfrac{T}{12}

, so

T = 12 \tan 35^\circ

T = 8.4024\ldots

Practise this →

3. Finding an angle

Opposite 3, adjacent 5: use the inverse

\displaystyle \tan\theta = \dfrac{3}{5}

e.g.

\displaystyle \theta = \tan^{-1} \dfrac{3}{5}

\displaystyle \theta = 31.0^\circ\text{ (1 d.p.)}

📖 Worked example

In the diagram, $\angle DEF = 90^\circ$ . Find $x$ , correct to 1 decimal place.

Step 1 of 3

Label the two known sides from the angle x°

From the angle x°: the side

8

cm is the opposite and the side

12.5

cm is the hypotenuse.

📖 Worked example

In the diagram, $\angle STU = 90^\circ$ , $SU = 15$ cm and $TU = 9$ cm. Write down the value of $\sin \angle UST$ , giving your answer as a fraction in its simplest form.

Step 1 of 2

Label the two sides that sin uses, from ∠UST

From

\angle UST

TU

is the opposite and

SU

is the hypotenuse.

📖 Worked example

In the diagram, $\angle KLM = 90^\circ$ , $KM = 20$ cm and $\sin \angle LKM = \dfrac{4}{5}$ . The point $N$ lies on $LM$ such that $LN : NM = 1 : 3$ . Find the length of $LN$ .

Step 1 of 2

The ratio tells you the sides in proportion: sin uses LM and KM

\dfrac{LM}{KM} = \dfrac{4}{5}

(from the sin ratio)

LM = \dfrac{4}{5} \times 20

= 16

📖 Worked example

Step 1 of 4

Label the two known sides from the angle x° at the foot of the ladder

From the angle x° at the foot of the ladder: the distance from the wall,

1.2

m, is the adjacent and the ladder,

4

m, is the hypotenuse.

Practise this →