Year 11 Maths - Specialist (Unit 1 and Unit 2) Trigonometry

Sine and cosine rules in two and three dimensions

Q: When do I use the sine rule versus the cosine rule?

Use the sine rule, a over sin A equals b over sin B equals c over sin C, whenever a side is paired with its opposite angle. Use the cosine rule, a squared equals b squared plus c squared minus 2bc cos A, when you know two sides and the angle between them, to find the third side, or all three sides, to find an angle.

Q: What is the cosine rule rearranged to find an angle?

From a squared equals b squared plus c squared minus 2bc cos A, make cos A the subject: cos A equals b squared plus c squared minus a squared, all over 2bc. Then A is the inverse cosine of that. A negative result means the angle is obtuse.

Q: How do I find the area of a triangle without its height?

Use area equals one half a b sin C, where a and b are two sides and C is the angle between them. For example two sides 12 and 9 with an included 40 degrees give one half times 12 times 9 times sin 40 degrees, about 34.7 square units.

Q: What is the ambiguous case of the sine rule?

When the sine rule is used to find an angle from two sides and a non-included angle, both an acute value B and its obtuse partner, 180 degrees minus B, have the same sine. If b sin A is less than a which is less than b, both give a valid triangle, so there are two solutions.

Q: How do I use these rules in three dimensions?

Find a single triangle inside the solid, often right angled, and redraw it flat. For a right pyramid of base side 10 and height 12, the right triangle T O M, with O M equals 5, gives the slant height T M equals the square root of 5 squared plus 12 squared, which is 13.

Q: How do I find the angle between two planes?

Locate the line where the planes meet, draw a perpendicular to that line in each plane from the same point, and measure the angle between them. For a pyramid face and its base this is the angle T M O, found from tan of angle T M O equals T O over O M.

20 practice questions 0 video lessons Theory + worked examples

Master the sine and cosine rules in two and three dimensions in Year 11 VCE Specialist Mathematics. These rules extend trigonometry from right-angled triangles to any triangle, using the sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), the cosine rule \(a^2=b^2+c^2-2bc\cos A\) and the area rule \(\tfrac12 ab\sin C\). It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Trigonometry topic of Unit 2.

You will learn to find unknown sides, angles and areas with the sine rule and cosine rule, handle the ambiguous case, and solve three-dimensional problems — pyramids, boxes and the angle between two planes — by isolating a well-chosen triangle, extending triangle trigonometry to real applications.

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Theory

The sine and cosine rules extend trigonometry from right-angled triangles to any triangle. In Year 11 Specialist Mathematics you use the sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), the cosine rule \(a^2=b^2+c^2-2bc\cos A\) and the area rule \(\text{Area}=\tfrac12 ab\sin C\) to find unknown sides, angles and areas, handle the ambiguous case, and solve three-dimensional problems — pyramids, boxes and the angle between two planes — through a well-chosen triangle.

Label a triangle so that side \(a\) is opposite angle \(A\), side \(b\) opposite \(B\) and side \(c\) opposite \(C\). With this convention two general rules relate the sides and angles of any triangle, not just right-angled ones.

The sine rule pairs each side with the sine of its opposite angle: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). Use it when you know a side and its opposite angle, plus one more side or angle — that is, two angles and a side (AAS/ASA), or two sides and a non-included angle.

The cosine rule \(a^2=b^2+c^2-2bc\cos A\) generalises Pythagoras' theorem (the extra term \(-2bc\cos A\) vanishes when \(A=90^\circ\)). Use it when you know two sides and the included angle (to find the third side) or all three sides (rearranged, to find an angle).

The area rule \(\text{Area}=\tfrac12 ab\sin C\) gives the area from two sides and the angle between them. The ambiguous case arises when the sine rule is used to find an angle from two sides and a non-included angle: because \(\sin\theta=\sin(180^\circ-\theta)\), there may be two possible triangles.

In three dimensions, the strategy is to spot a single triangle inside the solid — often a right-angled one — and apply Pythagoras or these rules to it. The angle between two planes is measured where a line in one plane meets the common edge at right angles, as with the angle between a pyramid's sloping face and its base.

The standard labelling: side \(a\) is opposite angle \(A\), \(b\) opposite \(B\), \(c\) opposite \(C\). Both rules pair a side with its opposite angle.

A 3-D problem solved through one triangle: the height \(TO\), the half-base \(OM\) and the slant \(TM\) form a right triangle. The angle \(\theta\) at \(M\) is the angle between the sloping face and the base.

In a box the space diagonal \(d\) satisfies \(d^2=l^2+w^2+h^2\) — Pythagoras applied twice, first across a face, then up to the opposite corner.

The sine rule pairs each side with the sine of its opposite angle:

\[ \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} \]

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

The cosine rule finds a side from two sides and the included angle:

\[ a^2=b^2+c^2-2bc\cos A \]

a^{2} = b^{2} + c^{2} - 2 b c \cos A

Rearranged, it finds an angle from all three sides:

\[ \cos A=\dfrac{b^2+c^2-a^2}{2bc} \]

The area from two sides and the angle between them:

\[ \text{Area}=\tfrac12\,ab\sin C \]

Area = \frac{1}{2} a b \sin C

In three dimensions, the space diagonal of a box with edges \(l,w,h\) comes from Pythagoras in 3-D:

\[ d^2=l^2+w^2+h^2 \]

Which rule? Reach for the sine rule when a side is paired with its opposite angle; reach for the cosine rule when you have two sides and the angle between them, or all three sides. If finding an angle from the sine rule with two sides and a non-included angle, check the ambiguous case — the obtuse partner \(180^\circ-B\) may also fit.

Choosing and applying a rule

Draw and label the triangle so each side sits opposite its angle (\(a\) opposite \(A\), and so on). Mark what you know.
Match the information to a rule. A side with its opposite angle \(\Rightarrow\) sine rule. Two sides and the included angle, or three sides \(\Rightarrow\) cosine rule. Two sides and the angle between them, for area \(\Rightarrow\) \(\tfrac12 ab\sin C\).
Substitute and solve. Rearrange first, then substitute the numbers; take \(\sqrt{\ }\) for a side, or \(\sin^{-1}\!/\cos^{-1}\) for an angle.
Check the ambiguous case. When the sine rule gives an angle from two sides and a non-included angle, test whether the obtuse value \(180^\circ-B\) also makes a valid triangle (it does when \(b\sin A < a < b\)).
In 3-D, isolate one triangle. Redraw the relevant right-angled or oblique triangle on its own — a height with a half-diagonal, a face with the base — then apply Pythagoras or a rule to that flat triangle.

For the angle between two planes (such as a pyramid's face and its base), find the line where they meet, drop a perpendicular to that edge in each plane, and measure the angle between those perpendiculars — for a right pyramid this is the angle \(\angle TMO\) in the right triangle \(TMO\).

Example 1 — Sine rule: find a side

In triangle \(ABC\), \(A=40^\circ\), \(B=60^\circ\) and side \(a=10\) (opposite \(A\)). Find side \(b\), to \(1\) decimal place.

Solution

A side is paired with its opposite angle, so use the sine rule, pairing \(b\) with \(a\):

\(\dfrac{b}{\sin B}\)	\(=\)	\(\dfrac{a}{\sin A}\)
\(b\)	\(=\)	\(\dfrac{a\,\sin B}{\sin A}\)
	\(=\)	\(\dfrac{10\,\sin 60^\circ}{\sin 40^\circ}\)
	\(=\)	\(\dfrac{10 \times 0.8660}{0.6428}\)
	\(=\)	\(13.47\ldots\)

Side \(b \approx 13.5\).

Example 2 — Cosine rule: find a side

In triangle \(ABC\), \(b=7\), \(c=9\) and the included angle \(A=55^\circ\). Find side \(a\), to \(1\) decimal place.

Solution

Two sides and the angle between them are given, so use the cosine rule:

\(a^2\)	\(=\)	\(b^2+c^2-2bc\cos A\)
	\(=\)	\(7^2+9^2-2(7)(9)\cos 55^\circ\)
	\(=\)	\(49+81-126 \times 0.5736\)
	\(=\)	\(130-72.27\)
	\(=\)	\(57.73\)

Take the square root:

\(a\)	\(=\)	\(\sqrt{57.73}\)
	\(=\)	\(7.60\ldots\)

Side \(a \approx 7.6\).

Example 3 — Cosine rule: find an angle

A triangle has sides \(a=8\), \(b=5\) and \(c=6\). Find the largest angle (angle \(A\), opposite the longest side), to \(1\) decimal place.

Solution

All three sides are known, so use the rearranged cosine rule for angle \(A\):

\(\cos A\)	\(=\)	\(\dfrac{b^2+c^2-a^2}{2bc}\)
	\(=\)	\(\dfrac{5^2+6^2-8^2}{2(5)(6)}\)
	\(=\)	\(\dfrac{25+36-64}{60}\)
	\(=\)	\(\dfrac{-3}{60}\)
	\(=\)	\(-0.05\)

A negative cosine means \(A\) is obtuse; take the inverse cosine:

\(A\)	\(=\)	\(\cos^{-1}(-0.05)\)
	\(=\)	\(92.9^\circ\)

The largest angle is \(A \approx 92.9^\circ\).

Example 4 — Area from two sides and the included angle

A triangle has sides \(a=12\) and \(b=9\) with an included angle \(C=40^\circ\). Find its area, to \(1\) decimal place.

Solution

Two sides and the angle between them are given, so use \(\text{Area}=\tfrac12 ab\sin C\):

\(\text{Area}\)	\(=\)	\(\tfrac12\,ab\sin C\)
	\(=\)	\(\tfrac12 \times 12 \times 9 \times \sin 40^\circ\)
	\(=\)	\(54 \times 0.6428\)
	\(=\)	\(34.71\ldots\)

The area is \(\approx 34.7\) square units.

Example 5 — The ambiguous case

In triangle \(ABC\), \(a=9\), \(b=12\) and \(A=40^\circ\). Find both possible values of angle \(B\), to \(1\) decimal place.

Solution

Two sides and a non-included angle: use the sine rule for \(\sin B\):

\(\dfrac{\sin B}{b}\)	\(=\)	\(\dfrac{\sin A}{a}\)
\(\sin B\)	\(=\)	\(\dfrac{12\,\sin 40^\circ}{9}\)
	\(=\)	\(0.8571\)

The acute solution, then its obtuse partner (supplement):

\(B\)	\(=\)	\(\sin^{-1}(0.8571) = 59.0^\circ\)
\(B'\)	\(=\)	\(180^\circ - 59.0^\circ = 121.0^\circ\)

Both are valid because \(b\sin A < a < b\) — that is \(7.71 < 9 < 12\) — so two triangles fit the data.

Angle \(B \approx 59.0^\circ\) or \(B \approx 121.0^\circ\).

Example 6 — Three dimensions: pyramid

A right pyramid has a square base \(ABCD\) of side \(10\) and apex \(T\) that sits \(12\) directly above the centre \(O\). \(M\) is the midpoint of edge \(AB\). Find the slant height \(TM\) and the angle between a sloping face and the base, to \(1\) decimal place.

Solution

Isolate the right triangle \(TOM\): \(O\) is directly below \(T\), and \(OM\) is half the base \((OM=5)\). For the slant height use Pythagoras:

\(TM^2\)	\(=\)	\(OM^2+TO^2\)
	\(=\)	\(5^2+12^2\)
	\(=\)	\(25+144 = 169\)
\(TM\)	\(=\)	\(\sqrt{169} = 13\)

The angle between the sloping face and the base is \(\angle TMO\) in the same right triangle:

\(\tan(\angle TMO)\)	\(=\)	\(\dfrac{TO}{OM}\)
	\(=\)	\(\dfrac{12}{5} = 2.4\)
\(\angle TMO\)	\(=\)	\(\tan^{-1}(2.4) = 67.4^\circ\)

The slant height is \(TM = 13\); the face meets the base at \(\approx 67.4^\circ\).

Common pitfalls

Using the wrong rule. The sine rule needs a side paired with its opposite angle. With two sides and the angle between them (or all three sides) the sine rule cannot start — use the cosine rule.

Forgetting the ambiguous case. When the sine rule gives an angle from two sides and a non-included angle, the obtuse partner \(180^\circ-B\) may also be valid. Always check whether a second triangle fits before committing to the acute answer.

Rounding too early. Keep full calculator accuracy through the working and round only the final side or angle. Rounding \(\cos A\) or an intermediate side first can shift the answer by a degree or more.

Missing the minus sign in the cosine rule. It is \(a^2=b^2+c^2-2bc\cos A\). When the angle is obtuse, \(\cos A\) is negative, so \(-2bc\cos A\) becomes a positive addition — the third side is longer, as expected.

Working in the sloping face in 3-D. The angle between two planes is measured along a perpendicular to their common edge, not along a slant edge. In a pyramid it is \(\angle TMO\) (through the midpoint \(M\)), not the angle at a corner.

Frequently asked questions

When do I use the sine rule versus the cosine rule?

Use the sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) whenever a side is paired with its opposite angle. Use the cosine rule \(a^2=b^2+c^2-2bc\cos A\) when you know two sides and the angle between them (to find the third side) or all three sides (to find an angle).

What is the cosine rule rearranged to find an angle?

From \(a^2=b^2+c^2-2bc\cos A\), make \(\cos A\) the subject: \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\). Then \(A=\cos^{-1}\!\left(\dfrac{b^2+c^2-a^2}{2bc}\right)\). A negative result means the angle is obtuse.

How do I find the area of a triangle without its height?

Use \(\text{Area}=\tfrac12 ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the angle between them. For example two sides \(12\) and \(9\) with an included \(40^\circ\) give \(\tfrac12(12)(9)\sin 40^\circ\approx 34.7\) square units.

What is the ambiguous case of the sine rule?

When the sine rule is used to find an angle from two sides and a non-included angle, both an acute value \(B\) and its obtuse partner \(180^\circ-B\) have the same sine. If \(b\sin A < a < b\) both give a valid triangle, so there are two solutions.

How do I use these rules in three dimensions?

Find a single triangle inside the solid — often right-angled — and redraw it flat. For a right pyramid of base side \(10\) and height \(12\), the right triangle \(TOM\) (with \(OM=5\)) gives the slant height \(TM=\sqrt{5^2+12^2}=13\).

How do I find the angle between two planes?

Locate the line where the planes meet, draw a perpendicular to that line in each plane from the same point, and measure the angle between them. For a pyramid face and its base this is \(\angle TMO\), found from \(\tan(\angle TMO)=\dfrac{TO}{OM}\).

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Sine and cosine rules in two and three dimensions

📖 Theory

Choosing and applying a rule

Common pitfalls

Frequently asked questions

Theory