Year 11 General Applications Of Trigonometry

The Sine Rule

Q: What is the sine rule?

The sine rule states that in any triangle, a over sin A equals b over sin B equals c over sin C. Each side, divided by the sine of its opposite angle, gives the same value (which equals the diameter of the circumscribed circle).

Q: When do I use the sine rule?

Use the sine rule when you have a matched pair: at least one side and the angle directly opposite it, plus one more piece of information (another angle or another side). If you only have three sides (SSS) or two sides plus the included angle (SAS), use the cosine rule instead.

Q: How do I rearrange the sine rule to find an angle?

Flip every fraction so the sines are on top: sin A over a equals sin B over b equals sin C over c. Then solve for the unknown sine and apply sin inverse. Writing the rule with sines on top saves one rearrangement step when the unknown is an angle.

Q: What does AAS mean in triangle problems?

AAS stands for Angle-Angle-Side — you're given two angles and one side of a triangle. The third angle is found from A plus B plus C equals 180 degrees, and the remaining sides via the sine rule. AAS data always defines a unique triangle (no ambiguous case).

13 practice questions 2 video lessons Theory + worked examples

Theory

The sine rule works for any triangle: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Use it whenever you have a matched pair (one side and the angle opposite it), plus one more piece of information.

The sine rule relates each side of any triangle to the sine of the angle opposite that side:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

It works for any triangle, not just right-angled ones. The convention: each side is labelled with the same letter as the opposite vertex, but in lowercase. So vertex $A$ is opposite side $a$ , $B$ is opposite $b$ , and $C$ is opposite $c$ .

You need a matched pair (one side and the angle directly opposite it) to use the sine rule. Two situations come up:

AAS — two angles and one side. Use $A + B + C = 180^{\circ}$ to find the third angle, then the sine rule for unknown sides.
SSA — two sides and an angle opposite one of them. Use the sine rule to find a second angle. Beware the ambiguous case (see Pitfalls).

Side

a

is opposite vertex

A

b

opposite

B

c

opposite

C

Ambiguous case: same data, two valid triangles (acute and obtuse)

The sine rule

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Finding a side

Use the formula the right way up:

a = \frac{b \sin A}{\sin B}

a = \frac{b \sin A}{\sin B}

Finding an angle

Flip the rule (take the reciprocal of every fraction) so the unknown sine is on top:

\frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \sin A = \frac{a \sin B}{b}

\sin A = \frac{a \sin B}{b}

          The ambiguous case (SSA): when you're given two sides and a non-included angle, there may be two valid triangles. After computing sin⁡A, the angle could be the acute solution OR its obtuse complement 180∘−A. Check both — only one might fit (angles must sum to less than 180∘).
        

How to use the sine rule

Label the triangle so each side has the lowercase of its opposite vertex ( $a$ opposite $A$ , etc).
Check you have a matched pair — at least one side and its opposite angle. If not, use the cosine rule instead.
Choose the form: side on top when finding a side; sine on top when finding an angle.
Solve — and if SSA, check whether the obtuse solution $180^{\circ} - A$ also fits.

Example 1 — AAS, find a side

In triangle

A B C

∠ A = 65^{\circ}

∠ B = 48^{\circ}

, and

a = 12

cm. Find

b

(1 dp).

Solution

Sketch and label the triangle:

$\frac{b}{\sin B}$	$=$	$\frac{a}{\sin A}$
$\frac{b}{\sin 48^{\circ}}$	$=$	$\frac{12}{\sin 65^{\circ}}$
$b$	$=$	$\frac{12 \sin 48^{\circ}}{\sin 65^{\circ}}$
$b$	$\approx$	$9.8 cm$

b = \frac{12 \sin 48 °}{\sin 65 °} \approx 9.8 cm

Example 2 — SSA, find an angle

In triangle

P Q R

p = 9

q = 7

, and

∠ P = 75^{\circ}

. Find

∠ Q

(nearest degree).

Solution

Use the sine rule with sines on top:

$\frac{\sin Q}{q}$	$=$	$\frac{\sin P}{p}$
$\sin Q$	$=$	$\frac{7 \sin 75^{\circ}}{9}$
$\sin Q$	$\approx$	$0.7510$
$Q$	$\approx$	$49^{\circ}$

Q = \sin^{- 1} (0.7510) \approx 49 °

Note: since $p > q$ and $∠ P$ is the largest angle here, the obtuse solution $131^{\circ}$ doesn't fit (it would make the angles sum to more than $180^{\circ}$ ).

Example 3 — AAS, full triangle

In triangle

A B C

∠ A = 52^{\circ}

∠ B = 71^{\circ}

a = 8

cm. Find

∠ C

and side

c

(1 dp).

Solution

Find $∠ C$ from the angle sum:

$∠ C$	$=$	$180^{\circ} - 52^{\circ} - 71^{\circ}$
$∠ C$	$=$	$57^{\circ}$

Then use the sine rule for side $c$ :

$c$	$=$	$\frac{a \sin C}{\sin A}$
$c$	$=$	$\frac{8 \sin 57^{\circ}}{\sin 52^{\circ}}$
$c$	$\approx$	$8.5 cm$

c = \frac{8 \sin 57 °}{\sin 52 °} \approx 8.5 cm

Example 4 — Ambiguous case

In triangle

A B C

∠ A = 40^{\circ}

A B = 10

B C = 7

. Find both possible values of

∠ C

(nearest minute).

Solution

$∠ A = 40^{\circ}$ is opposite $B C = 7$ . $A B = 10$ is opposite $∠ C$ . Both the acute and obtuse solutions for $C$ fit because $A B > B C$ :

$\sin C$	$=$	$\frac{10 \sin 40^{\circ}}{7}$
$\sin C$	$\approx$	$0.9183$
$C$	$\approx$	$66^{\circ} 40^{'}$
or $C$	$\approx$	$113^{\circ} 20^{'}$

C \approx 66 ° 40' or 113 ° 20'

Common pitfalls

Wrong tool for the job. The sine rule will NOT work with three sides (use the cosine rule) or two sides and the included angle (also cosine rule). You need a matched side-and-opposite-angle pair.

Forgetting the obtuse solution. When finding an angle from SSA data, the calculator gives only the acute solution. The obtuse

180^{\circ} - A

may also fit — always check both before deciding.

Putting the unknown on the bottom of the fraction. If you're finding an angle, write the formula with sines on top (

\frac{\sin A}{a} = \frac{\sin B}{b}

) — it saves an extra algebra step.

Mixing up sides and opposite angles. Each fraction in the rule must pair a side with the angle opposite it, not an adjacent angle. Double-check your labelling before substituting.

Frequently asked questions

What is the sine rule?

The sine rule states that in any triangle, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Each side, divided by the sine of its opposite angle, gives the same value (which equals the diameter of the circumscribed circle).

When do I use the sine rule?

Use the sine rule when you have a matched pair: at least one side and the angle directly opposite it, plus one more piece of information (another angle or another side). If you only have three sides (SSS) or two sides plus the included angle (SAS), use the cosine rule instead.

What is the ambiguous case?

When you're given two sides and a non-included angle (SSA), the data may fit two different triangles. After finding $\sin$ of an angle, the angle could be the acute value from $\sin^{- 1}$ or its obtuse complement $180^{\circ} - that value$ . Always check whether the obtuse solution also fits (angles must sum to less than $180^{\circ}$ ).

Does the sine rule work for right-angled triangles?

Yes, but SOH-CAH-TOA is usually simpler for right-angled triangles. The sine rule works for any triangle and reduces correctly when one angle is $90^{\circ}$ , so you can use it if you want — it just isn't usually the fastest method.

How do I rearrange the sine rule to find an angle?

Flip every fraction so the sines are on top: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ . Then solve for the unknown sine and apply $\sin^{- 1}$ . Writing the rule with sines on top one step saves rearrangement when the unknown is an angle.

What does AAS mean in triangle problems?

AAS stands for "Angle-Angle-Side" — you're given two angles and one side of a triangle. The third angle is found from $A + B + C = 180^{\circ}$ , and the remaining sides via the sine rule. AAS data always defines a unique triangle (no ambiguous case).