Trigonometry and Pythagoras in 2D and 3D

Turning a complicated figure into right-angled triangles

Pythagoras’ theorem and the ratios sine, cosine and tangent apply directly to right-angled triangles. A question may instead show a rectangle, prism, cuboid or another general two- or three-dimensional figure. The main challenge is therefore not usually choosing a formula immediately: it is identifying or constructing the right-angled triangle that contains the required length or angle.

Useful lines include a diagonal across a face, a perpendicular height and the perpendicular projection of a sloping line onto a base. Mark the right angle justified by the geometry. For example, adjacent edges of a rectangular face meet at $90^\circ$.

Multi-step problems in two dimensions

A general 2D figure can often be divided into right-angled triangles. Drawing a diagonal across a rectangle produces a triangle whose legs are the rectangle’s length and width. Dropping a perpendicular from the apex of an isosceles triangle to its base produces two congruent right-angled triangles, so each contains half of the base.

Once the useful triangle is visible, use the information it contains:

• use Pythagoras when two side lengths are known and a third is required;
• use sine, cosine or tangent when an angle and a side are involved;
• use an inverse trigonometric function when an angle is required from two sides.

The original figure need not itself be a triangle. What matters is that the calculation takes place within a correctly identified right-angled triangle.

Building a route through a 3D solid

Three-dimensional questions are assessed at Higher tier. Although the diagram is 3D, the calculation is usually a sequence of 2D calculations in different planes. A length found in one face can become a side of a second right-angled triangle.

Consider a cuboid with base dimensions $8\text{ cm}$ by $5\text{ cm}$ and height $4\text{ cm}$. First find the base diagonal $d$:

$$d^2=8^2+5^2=89$$

so $d=\sqrt{89}\text{ cm}$. The base diagonal and the vertical height then form another right-angled triangle whose hypotenuse is the space diagonal $D$:

$$D^2=(\sqrt{89})^2+4^2=105$$

Therefore $D=\sqrt{105}\text{ cm}\approx10.2\text{ cm}$. This is sometimes summarised for a cuboid as

$$D=\sqrt{l^2+w^2+h^2}$$

but the two-triangle method explains why the result works and can be adapted to less regular solids.

Finding an angle between a line and a plane

The angle between a line and a plane is never measured against an arbitrary edge of the plane. It is the acute angle between the line and its perpendicular projection onto that plane.

Suppose a sloping line rises $5\text{ cm}$ above a rectangular base whose perpendicular horizontal dimensions are $9\text{ cm}$ and $4\text{ cm}$. Its projection across the base has length

$$\sqrt{9^2+4^2}=\sqrt{97}\text{ cm}$$

The sloping line, its projection and the vertical rise now form a right-angled triangle. If $\theta$ is the angle between the line and the base plane, then

$$\tan\theta=\frac{5}{\sqrt{97}}$$

so

$$\theta=\tan^{-1}\left(\frac{5}{\sqrt{97}}\right)\approx26.9^\circ$$

The calculation has two stages because the necessary horizontal projection was not labelled directly on the original solid.

Choosing and ordering the calculations

Work backwards from the requested quantity. Locate a right-angled triangle containing it, then ask whether that triangle already has enough known information. If not, identify another right-angled triangle that supplies the missing side.

A reliable written method is:

1. mark or draw the required diagonal, height or projection;
2. sketch the first right-angled triangle in 2D and label known sides;
3. calculate the shared length;
4. transfer that length to the second right-angled triangle;
5. apply Pythagoras or the appropriate trigonometric ratio;
6. round only the final answer and include the correct unit.

Lengths use linear units such as cm. Angles are normally given in degrees. Retaining a surd such as $\sqrt{89}$ between stages avoids an error caused by premature rounding.

Core idea

Break a 2D or 3D figure into right-angled triangles. A result from one triangle may become a side of the next.

Formulae

For hypotenuse $c$:

$$a^2+b^2=c^2$$

Relative to angle $\theta$:

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$

Use an inverse ratio to find an angle.

3D route

For a cuboid, find a face diagonal first, then combine it with the perpendicular third dimension. Equivalently,

$$D=\sqrt{l^2+w^2+h^2}$$

Keep intermediate values exact where possible. Three-dimensional problems are Higher tier.

Line and plane

The angle between a line and a plane is the acute angle between the line and its perpendicular projection onto the plane. Find the projection first, then use the right-angled triangle containing the projection and the line.