Sec 3 · G2 (N(A))

Further trigonometry

Practise further trigonometry with step-by-step help. Free account.

1. Sine and cosine of obtuse angles

Rule	Example
$\sin(180^\circ - x) = \sin x$	$\sin 145^\circ = \sin 35^\circ$
$\cos(180^\circ - x) = -\cos x$	$\cos 140^\circ = -\cos 40^\circ$

Sine of an obtuse angle

\displaystyle \sin 145^\circ

e.g.

\displaystyle \sin 145^\circ

\displaystyle = \sin(180^\circ - 145^\circ)

\displaystyle = \sin 35^\circ

\displaystyle = 0.574\text{ (3 s.f.)}

Cosine of an obtuse angle

\displaystyle \cos 140^\circ

e.g.

\displaystyle \cos 140^\circ

\displaystyle = -\cos(180^\circ - 140^\circ)

\displaystyle = -\cos 40^\circ

\displaystyle = -0.766\text{ (3 s.f.)}

Two angles with the same sine

\displaystyle \sin x = 0.4, \quad 0^\circ < x < 180^\circ

acute: $x = \sin^{-1}(0.4)$ , so $x = 23.6^\circ$
obtuse: $x = 180^\circ - 23.6^\circ$ , so $x = 156.4^\circ$

📖 Worked example

Given that $\cos 35^\circ = 0.819$ , write down the value of $\cos 145^\circ$ .

Step 1 of 1

Use the value you are given

\cos 145^\circ = -\cos(180^\circ - 145^\circ)

\phantom{\cos 145^\circ} = -\cos 35^\circ

\phantom{\cos 145^\circ} = -0.819

📖 Worked example

Given that $\sin x = 0.4$ and $0^\circ < x < 180^\circ$ , find the two possible values of $x$ , correct to 1 decimal place.

Step 1 of 3

The calculator gives the acute angle

\sin x = 0.4

x = \sin^{-1}(0.4)

x = 23.5781\ldots^\circ

\phantom{x} = 23.6^\circ

(1 d.p.)

Practise this →

2. Area of a triangle

Find	Formula
Area	$\dfrac{1}{2}ab\sin C$
Angle	$\sin C = \dfrac{2 \times \text{Area}}{ab}$
Side	$b = \dfrac{2 \times \text{Area}}{a\sin C}$

a = 7 cm, b = 7 cm, C = 134°

\displaystyle \dfrac{1}{2}ab\sin C

e.g.

\displaystyle \dfrac{1}{2}(7)(7)\sin 134^\circ

\displaystyle = 17.6\text{ (3 s.f.)}

Area $= 17.6$ cm²

Area 15 cm², a = 7 cm, b = 8 cm, C obtuse

\displaystyle \sin C = \dfrac{2(15)}{7 \times 8}

acute angle: $\sin^{-1}\left(\dfrac{30}{56}\right) = 32.39^\circ$
$C = 180^\circ - 32.39^\circ$ , so $C = 147.6^\circ$

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $CA = 7$ cm and angle $ACB = 134^\circ$ . Find the area of the triangle, correct to 3 significant figures.

Step 1 of 2

Two sides and the angle between them

\text{Area} = \dfrac{1}{2}ab \sin C

a = 7

b = 7

C = 134^\circ

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $CA = 8$ cm and the area is $15$ cm². Given that angle $ACB$ is obtuse, find angle $ACB$ , correct to 1 decimal place.

Step 1 of 3

Substitute into the area formula

\text{Area} = \dfrac{1}{2}ab \sin C

15 = \dfrac{1}{2} \times 7 \times 8 \times \sin \theta

Practise this →

3. Sine rule

Find	Formula
a side	$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
an angle	$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

A side: angles 48° and 36° opposite 12 cm and x cm

\displaystyle \dfrac{x}{\sin 36^\circ} = \dfrac{12}{\sin 48^\circ}

e.g.

\displaystyle \dfrac{x}{\sin 36^\circ} = \dfrac{12}{\sin 48^\circ}

\displaystyle x = \dfrac{12\sin 36^\circ}{\sin 48^\circ}

\displaystyle x = 9.49\text{ (3 s.f.)}

An angle: 10 cm opposite θ, 7 cm opposite 38°, θ obtuse

\displaystyle \dfrac{\sin \theta}{10} = \dfrac{\sin 38^\circ}{7}

$\sin \theta = \dfrac{10\sin 38^\circ}{7}$
acute angle: $61.58^\circ$
$\theta = 180^\circ - 61.58^\circ$ , so $\theta = 118.4^\circ$

📖 Worked example

In triangle $ABC$ , angle $BAC = 48^\circ$ , angle $ACB = 96^\circ$ and $BC = 12$ cm. Find the length of $AC$ , correct to 3 significant figures.

Step 1 of 5

Find the angle opposite the side you want

Angle

ABC = 180^\circ - 48^\circ - 96^\circ

ABC = 36^\circ

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $AC = 10$ cm and angle $BAC = 38^\circ$ . Given that angle $ABC$ is obtuse, find angle $ABC$ , correct to 1 decimal place.

Step 1 of 5

Which rule links the sides and angles you know with the one you want?

\text{Sine rule}

Practise this →

4. Cosine rule

Find	Formula
a side	$a^{2} = b^{2} + c^{2} - 2bc\cos A$
an angle	$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$

You know	Use
two angles and a side	sine rule
two sides and an angle opposite one of them	sine rule
two sides and the angle between them	cosine rule
three sides	cosine rule

A side: 6 cm and 14 cm with 125° between them

\displaystyle x^{2} = 6^{2} + 14^{2} - 2(6)(14)\cos 125^\circ

e.g.

\displaystyle x = \sqrt{6^{2} + 14^{2} - 2(6)(14)\cos 125^\circ}

\displaystyle x = 18.1\text{ (3 s.f.)}

An angle: sides 7 cm, 8 cm, 14 cm; θ opposite 7 cm

\displaystyle \cos \theta = \dfrac{8^{2} + 14^{2} - 7^{2}}{2(8)(14)}

$\theta = \cos^{-1}\left(\dfrac{211}{224}\right)$ , so $\theta = 19.6^\circ$

📖 Worked example

In triangle $ABC$ , $AB = 14$ cm, $AC = 6$ cm and angle $BAC = 125^\circ$ . Find the length of $BC$ , correct to 3 significant figures.

Step 1 of 4

Which rule links the sides and angles you know with the one you want?

\text{Cosine rule}

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $AC = 8$ cm and $AB = 14$ cm. Find angle $BAC$ , correct to 1 decimal place.

Step 1 of 4

Which rule links the sides and angles you know with the one you want?

\text{Cosine rule}

Practise this →

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Sec 3 · G2 (N(A))

Further trigonometry

Practise further trigonometry with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Sine and cosine of obtuse angles

Rule	Example
$\sin(180^\circ - x) = \sin x$	$\sin 145^\circ = \sin 35^\circ$
$\cos(180^\circ - x) = -\cos x$	$\cos 140^\circ = -\cos 40^\circ$

Sine of an obtuse angle

\displaystyle \sin 145^\circ

e.g.

\displaystyle \sin 145^\circ

\displaystyle = \sin(180^\circ - 145^\circ)

\displaystyle = \sin 35^\circ

\displaystyle = 0.574\text{ (3 s.f.)}

Cosine of an obtuse angle

\displaystyle \cos 140^\circ

e.g.

\displaystyle \cos 140^\circ

\displaystyle = -\cos(180^\circ - 140^\circ)

\displaystyle = -\cos 40^\circ

\displaystyle = -0.766\text{ (3 s.f.)}

Two angles with the same sine

\displaystyle \sin x = 0.4, \quad 0^\circ < x < 180^\circ

acute: $x = \sin^{-1}(0.4)$ , so $x = 23.6^\circ$
obtuse: $x = 180^\circ - 23.6^\circ$ , so $x = 156.4^\circ$

📖 Worked example

Given that $\cos 35^\circ = 0.819$ , write down the value of $\cos 145^\circ$ .

Step 1 of 1

Use the value you are given

\cos 145^\circ = -\cos(180^\circ - 145^\circ)

\phantom{\cos 145^\circ} = -\cos 35^\circ

\phantom{\cos 145^\circ} = -0.819

📖 Worked example

Given that $\sin x = 0.4$ and $0^\circ < x < 180^\circ$ , find the two possible values of $x$ , correct to 1 decimal place.

Step 1 of 3

The calculator gives the acute angle

\sin x = 0.4

x = \sin^{-1}(0.4)

x = 23.5781\ldots^\circ

\phantom{x} = 23.6^\circ

(1 d.p.)

Practise this →

2. Area of a triangle

Find	Formula
Area	$\dfrac{1}{2}ab\sin C$
Angle	$\sin C = \dfrac{2 \times \text{Area}}{ab}$
Side	$b = \dfrac{2 \times \text{Area}}{a\sin C}$

a = 7 cm, b = 7 cm, C = 134°

\displaystyle \dfrac{1}{2}ab\sin C

e.g.

\displaystyle \dfrac{1}{2}(7)(7)\sin 134^\circ

\displaystyle = 17.6\text{ (3 s.f.)}

Area $= 17.6$ cm²

Area 15 cm², a = 7 cm, b = 8 cm, C obtuse

\displaystyle \sin C = \dfrac{2(15)}{7 \times 8}

acute angle: $\sin^{-1}\left(\dfrac{30}{56}\right) = 32.39^\circ$
$C = 180^\circ - 32.39^\circ$ , so $C = 147.6^\circ$

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $CA = 7$ cm and angle $ACB = 134^\circ$ . Find the area of the triangle, correct to 3 significant figures.

Step 1 of 2

Two sides and the angle between them

\text{Area} = \dfrac{1}{2}ab \sin C

a = 7

b = 7

C = 134^\circ

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $CA = 8$ cm and the area is $15$ cm². Given that angle $ACB$ is obtuse, find angle $ACB$ , correct to 1 decimal place.

Step 1 of 3

Substitute into the area formula

\text{Area} = \dfrac{1}{2}ab \sin C

15 = \dfrac{1}{2} \times 7 \times 8 \times \sin \theta

Practise this →

3. Sine rule

Find	Formula
a side	$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
an angle	$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

A side: angles 48° and 36° opposite 12 cm and x cm

\displaystyle \dfrac{x}{\sin 36^\circ} = \dfrac{12}{\sin 48^\circ}

e.g.

\displaystyle \dfrac{x}{\sin 36^\circ} = \dfrac{12}{\sin 48^\circ}

\displaystyle x = \dfrac{12\sin 36^\circ}{\sin 48^\circ}

\displaystyle x = 9.49\text{ (3 s.f.)}

An angle: 10 cm opposite θ, 7 cm opposite 38°, θ obtuse

\displaystyle \dfrac{\sin \theta}{10} = \dfrac{\sin 38^\circ}{7}

$\sin \theta = \dfrac{10\sin 38^\circ}{7}$
acute angle: $61.58^\circ$
$\theta = 180^\circ - 61.58^\circ$ , so $\theta = 118.4^\circ$

📖 Worked example

In triangle $ABC$ , angle $BAC = 48^\circ$ , angle $ACB = 96^\circ$ and $BC = 12$ cm. Find the length of $AC$ , correct to 3 significant figures.

Step 1 of 5

Find the angle opposite the side you want

Angle

ABC = 180^\circ - 48^\circ - 96^\circ

ABC = 36^\circ

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $AC = 10$ cm and angle $BAC = 38^\circ$ . Given that angle $ABC$ is obtuse, find angle $ABC$ , correct to 1 decimal place.

Step 1 of 5

Which rule links the sides and angles you know with the one you want?

\text{Sine rule}

Practise this →

4. Cosine rule

Find	Formula
a side	$a^{2} = b^{2} + c^{2} - 2bc\cos A$
an angle	$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$

You know	Use
two angles and a side	sine rule
two sides and an angle opposite one of them	sine rule
two sides and the angle between them	cosine rule
three sides	cosine rule

A side: 6 cm and 14 cm with 125° between them

\displaystyle x^{2} = 6^{2} + 14^{2} - 2(6)(14)\cos 125^\circ

e.g.

\displaystyle x = \sqrt{6^{2} + 14^{2} - 2(6)(14)\cos 125^\circ}

\displaystyle x = 18.1\text{ (3 s.f.)}

An angle: sides 7 cm, 8 cm, 14 cm; θ opposite 7 cm

\displaystyle \cos \theta = \dfrac{8^{2} + 14^{2} - 7^{2}}{2(8)(14)}

$\theta = \cos^{-1}\left(\dfrac{211}{224}\right)$ , so $\theta = 19.6^\circ$

📖 Worked example

In triangle $ABC$ , $AB = 14$ cm, $AC = 6$ cm and angle $BAC = 125^\circ$ . Find the length of $BC$ , correct to 3 significant figures.

Step 1 of 4

Which rule links the sides and angles you know with the one you want?

\text{Cosine rule}

📖 Worked example

In triangle $ABC$ , $BC = 7$ cm, $AC = 8$ cm and $AB = 14$ cm. Find angle $BAC$ , correct to 1 decimal place.

Step 1 of 4

Which rule links the sides and angles you know with the one you want?

\text{Cosine rule}

Practise this →

Try a question: Further trigonometry

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