Year 12 Methods (Unit 3 & 4) Trigonometry using the sine and cosine rules

Problems in three dimensions

Q: How do you solve a trigonometry problem in three dimensions?

Draw the solid, then find the flat two-dimensional triangle that contains the length or angle you want. Work only with that triangle, using Pythagoras, right-triangle trigonometry, or the sine and cosine rules. Often you must first find a length such as a base diagonal in one triangle before using it in a second triangle.

Q: What is the space diagonal of a rectangular box?

The space diagonal joins a corner to the opposite corner through the inside of the box. For a box with length l, width w and height h it has length the square root of l squared plus w squared plus h squared. You get it by finding the base diagonal first, then using it as one leg of a right triangle with the height.

Q: What is the angle between a line and a plane?

It is the angle between the line and its shadow (projection) on the plane. To find the angle a space diagonal makes with the base of a box, use the right triangle whose horizontal leg is the base diagonal and whose vertical leg is the height, then take the inverse tangent of height divided by base diagonal.

Q: When do you use the sine rule or the cosine rule in a 3D problem?

Once you have isolated a two-dimensional triangle that is not right-angled. Use the cosine rule when you know two sides and the included angle, or all three sides. Use the sine rule when you know an angle and its opposite side plus one more angle or side.

Q: How do angles of elevation and depression work in three dimensions?

An angle of elevation is measured up from the horizontal to a higher point, and depression is measured down from the horizontal. In 3D you usually solve one vertical right triangle to get a height, then a horizontal triangle in the ground plane to get a distance, linking the two triangles by a shared side.

Q: Why must you draw the correct triangle first?

A common mistake is to use an edge of the solid instead of a diagonal, or a face that does not contain the required angle. Sketching the specific right triangle, and labelling which side is opposite, adjacent or the hypotenuse, makes sure the correct sides go into Pythagoras or the trig ratios.

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In Year 12 Mathematical Methods (Queensland, QCAA), a three-dimensional problem is solved by finding the flat two-dimensional triangle inside the solid that contains the length or angle you want, then applying Pythagoras (including the space diagonal of a box), right-triangle trigonometry, and the sine and cosine rules and area formula — extended to the angle between a line and a plane and to angles of elevation and depression solved with two linked triangles.

A problem in three dimensions asks for a length or an angle inside a solid such as a box, a pyramid, a wedge or a leaning pole. The key skill is to reduce it to two dimensions: locate the flat triangle that contains the quantity you want, redraw that triangle on its own, and solve it.

The tools are the ones from plane trigonometry:

• Pythagoras for right triangles — used twice to reach the space diagonal of a rectangular box (first the base diagonal, then the diagonal through the box).

• Right-triangle trigonometry (SOH–CAH–TOA) for a length or angle in a right triangle.

• The sine rule and cosine rule, and the area formula \(\tfrac{1}{2}ab\sin C\), for a triangle that is not right-angled.

The angle between a line and a plane is the angle between the line and its projection (its shadow) onto the plane. Angles of elevation and depression are measured from the horizontal — up for elevation, down for depression.

Key idea. Every 3D problem becomes a 2D one: sketch the solid, isolate the right (or non-right) triangle that holds the answer, label its sides, and solve that single triangle.

Space diagonal \(AG=\sqrt{l^{2}+w^{2}+h^{2}}\): find base diagonal \(AC\) first, then \(\triangle ACG\)

Angle between line and plane: \(\tan\theta=\dfrac{\text{height}}{\text{base diagonal}}\)

Pythagoras and the space diagonal of a box (length \(l\), width \(w\), height \(h\)):

\[c^{2}=a^{2}+b^{2},\qquad d=\sqrt{l^{2}+w^{2}+h^{2}}\]

d = \sqrt{l^{2} + w^{2} + h^{2}}

Right-triangle ratios (SOH–CAH–TOA):

\[\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\quad \cos\theta=\dfrac{\text{adj}}{\text{hyp}},\quad \tan\theta=\dfrac{\text{opp}}{\text{adj}}\]

The sine rule, cosine rule and area formula for any triangle:

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},\qquad c^{2}=a^{2}+b^{2}-2ab\cos C,\qquad \text{Area}=\dfrac{1}{2}ab\sin C\]

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

Angle between a line and a plane. It equals the angle between the line and its projection onto the plane. For the space diagonal of a box, \(\tan\theta=\dfrac{\text{height}}{\text{base diagonal}}\), so \(\theta=\tan^{-1}\!\left(\dfrac{\text{height}}{\text{base diagonal}}\right)\).

How to solve a problem in three dimensions

Draw the solid and mark every length and angle you are given.
Find the triangle that contains the length or angle you want, and redraw that flat triangle on its own.
Find any missing side you need first — often a face diagonal or base diagonal by Pythagoras — before you can solve the target triangle.
Choose the right tool. Right triangle: Pythagoras or SOH–CAH–TOA. Non-right triangle: cosine rule (two sides + included angle, or three sides), sine rule (an angle with its opposite side), or \(\tfrac{1}{2}ab\sin C\) for area.
Answer in context, with correct units and the required accuracy.

Two-triangle problems. For elevation and depression in 3D, solve one vertical right triangle to get a height, then a horizontal triangle in the ground plane for a distance, linking them by a shared side.

Example 1 — Space diagonal

A box is \(6\) cm long, \(3\) cm wide and \(2\) cm high. Find its space diagonal \(AG\).

Solution

Base diagonal \(AC\) first, then \(\triangle ACG\) (right angle at \(C\)).

\(AC^{2}\)	\(=\)	\(6^{2}+3^{2}=45\)
\(AG^{2}\)	\(=\)	\(AC^{2}+2^{2}=45+4=49\)
\(AG\)	\(=\)	\(\sqrt{49}=7\) cm

A G = 7

Example 2 — Line and plane

A box has base \(8\) cm by \(6\) cm and height \(6\) cm. Find the angle the space diagonal makes with the base.

Solution

Base diagonal \(=\sqrt{8^{2}+6^{2}}=10\); use \(\triangle ACG\).

\(\tan\theta\)	\(=\)	\(\dfrac{6}{10}\)
\(\theta\)	\(=\)	\(\tan^{-1}\!\left(\dfrac{6}{10}\right)\approx 31^{\circ}\)

θ \approx 31

Example 3 — Cosine rule on a face

A sloping face \(PAB\) of a pyramid has \(PA=8\) m, \(PB=15\) m and \(\angle APB=60^{\circ}\). Find \(AB\).

Solution

Two sides and the included angle ⇒ cosine rule.

\(AB^{2}\)	\(=\)	\(8^{2}+15^{2}-2(8)(15)\cos 60^{\circ}\)
	\(=\)	\(289-120=169\)
\(AB\)	\(=\)	\(13\) m

A B = 13

Example 4 — Elevation, two triangles

A mast \(OT\) is vertical. \(P\) is \(40\) m due west of \(O\); from \(P\) the elevation of \(T\) is \(45^{\circ}\). \(Q\) is due south with elevation \(30^{\circ}\). Find \(PQ\).

Solution

Height from \(\triangle OPT\), then \(OQ\), then Pythagoras.

\(OT\)	\(=\)	\(40\tan 45^{\circ}=40\)
\(OQ\)	\(=\)	\(\dfrac{40}{\tan 30^{\circ}}=40\sqrt{3}\)
\(PQ\)	\(=\)	\(\sqrt{40^{2}+(40\sqrt{3})^{2}}=80\) m

P Q = 80

Common pitfalls

Use the diagonal, not an edge. The horizontal leg of the space-diagonal triangle is the base diagonal, not a single edge of the box. Using an edge gives the wrong angle or length.

Pick the triangle that holds the angle. Only a face or plane that actually contains the required angle can be used; redraw that flat triangle before substituting into any ratio.

Right angle first. Pythagoras and SOH–CAH–TOA need a right angle. If the isolated triangle is not right-angled, switch to the sine or cosine rule instead.

Frequently asked questions

How do you solve a trigonometry problem in three dimensions?

Draw the solid, find the flat triangle that contains the length or angle you want, and solve just that triangle with Pythagoras, SOH–CAH–TOA, or the sine and cosine rules. Often you need a length from one triangle before solving a second.

What is the space diagonal of a rectangular box?

The diagonal from a corner to the opposite corner through the box: \(d=\sqrt{l^{2}+w^{2}+h^{2}}\). Find the base diagonal first, then use it as one leg of a right triangle with the height.

What is the angle between a line and a plane?

The angle between the line and its projection on the plane. For a space diagonal, \(\theta=\tan^{-1}\!\left(\dfrac{\text{height}}{\text{base diagonal}}\right)\).

When do you use the sine rule or the cosine rule in a 3D problem?

Once you have a non-right triangle: cosine rule for two sides and the included angle (or three sides), sine rule for an angle with its opposite side.

How do angles of elevation and depression work in three dimensions?

Elevation is measured up from the horizontal, depression down. Solve a vertical right triangle for a height, then a horizontal triangle for a distance, sharing one side.

Why must you draw the correct triangle first?

To avoid using an edge instead of a diagonal, or a face that does not contain the angle. Labelling opposite, adjacent and hypotenuse ensures the right sides enter each formula.

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