2D and 3D Problem Solving

Turn a spatial problem into triangles

A three-dimensional problem is usually solved through two-dimensional triangles. A triangle may lie on a face of a solid, across its interior or in a cross-section. The important question is not how the triangle looks in perspective, but which lengths and angles are known to be related.

Start with the length or angle you need. Identify a triangle containing it, then work backwards: does that triangle need a length you can find in another triangle? A base diagonal, for example, can become a side of a vertical triangle.

Choose a rule from the information available

For a right-angled triangle, Pythagoras’ theorem connects its three sides:

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

Here $c$ is the hypotenuse, opposite the right angle. To find a shorter side, rearrange by subtraction: $a=\sqrt{c^2-b^2}$.

The right-angled trigonometric ratios connect an angle $\theta$ with side lengths:

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

Opposite and adjacent are named relative to the angle being used; the adjacent side here is not the hypotenuse. Use an inverse trigonometric function to recover an angle from a ratio.

For a triangle without a right angle, label side $a$ opposite angle $A$, side $b$ opposite $B$ and side $c$ opposite $C$.

• Use the sine rule when you know an opposite side–angle pair and another relevant side or angle:

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

• Use the cosine rule when you know two sides and their included angle, or all three sides:

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

For an angle, this rearranges to

$$\cos A=\frac{b^2+c^2-a^2}{2bc}.$$

• For an area from two sides and their included angle, use

$$\text{Area}=\frac12 ab\sin C.$$

These rules also apply to triangles inside larger 2D shapes. For example, two sides of a triangle are $7$ cm and $9$ cm, with included angle $60^\circ$. The third side $a$ satisfies

$$a^2=7^2+9^2-2(7)(9)\cos60^\circ=67,$$

so its exact length is $\sqrt{67}$ cm. The cosine rule works because the given angle is between the two known sides.

When finding an angle using the sine rule, remember that an acute angle and its supplement have the same sine. If inverse sine gives $73.0^\circ$ but the required angle is obtuse, it is $180^\circ-73.0^\circ=107.0^\circ$. Use the triangle’s angle sum and the given conditions to decide which solution is valid.

Find a cuboid’s diagonal in two stages

Consider a cuboid with base dimensions $6$ cm by $8$ cm and vertical height $5$ cm. Label its base $ABCD$ in order, with $G$ vertically above $C$.

The base diagonal $AC$ lies in right-angled triangle $ABC$:

$$AC=\sqrt{6^2+8^2}=10\text{ cm}.$$

Now look at triangle $ACG$. Since $CG$ is perpendicular to the base plane, it is perpendicular to $AC$. The body diagonal is therefore

$$AG=\sqrt{AC^2+CG^2}=\sqrt{10^2+5^2}=5\sqrt5\text{ cm}.$$

The base diagonal is an intermediate length, not the body diagonal. Combining the two Pythagoras calculations gives $d^2=l^2+w^2+h^2$ for a cuboid with perpendicular dimensions .

Measure the angle between a line and a plane

The angle between a line and a plane is measured between the line and its perpendicular projection onto that plane. The projection is the line’s shadow when projected at right angles to the plane.

For the cuboid, $A$ already lies in the base, while $G$ projects vertically onto $C$. Therefore, the projection of $AG$ onto the base is $AC$, and the required angle is $\angle GAC$.

In triangle $ACG$, the vertical height is opposite this angle and the base diagonal is adjacent:

$$\tan\theta=\frac{CG}{AC}=\frac5{10}.$$

Thus $\theta=\tan^{-1}(1/2)\approx26.6^\circ$. This is the angle of the body diagonal above the base, not its angle with a base edge.

More generally, if a sloping line meets a plane at $A$, choose a point $P$ on the line and drop a perpendicular $PQ$ to the plane. Triangle $APQ$ is right-angled at $Q$, and the line–plane angle is $\angle PAQ$.

Measure the angle between two planes

Two intersecting planes meet along a straight line. To measure their dihedral angle, take a cross-section perpendicular to that line of intersection. The cross-section cuts each plane in a line; the angle between these two lines is the required plane–plane angle. An oblique cross-section generally gives a different angle.

Consider a triangular wedge with horizontal rectangular base $ABCD$, sloping rectangular faces $AEFD$ and $BEFC$, and triangular ends $ABE$ and $$. Its ridge is parallel to and . The two sloping rectangles are congruent, the base width is cm, and the maximum vertical height is cm.

The base and face $AEFD$ meet along $AD$. The triangular end $DCF$ is perpendicular to $AD$, so the angle between the planes is $\angle CDF$.

The congruent faces give $DF=CF$, making triangle $DCF$ isosceles. Its perpendicular height from $F$ meets $CD$ at its midpoint $M$. Consequently, cm and cm.

In right-angled triangle $DMF$,

$$\tan\angle FDM=\frac56,$$

so the angle between the sloping face and the base is $\tan^{-1}(5/6)\approx39.8^\circ$. The full base width would not be the adjacent side of this right-angled triangle.

Non-right-angled triangles inside solids

A useful cross-section need not be right-angled. For example, take a cone whose slant height is $2x$ cm and whose axial cross-section has apex angle $30^\circ$. A cross-section through its axis gives an isosceles triangle: the two sloping sides are each $2x$ cm and the side opposite the apex angle is the base diameter $d$.

Apply the cosine rule directly:

$$d^2=(2x)^2+(2x)^2-2(2x)(2x)\cos30^\circ.$$

Using $\cos30^\circ=\sqrt3/2$ gives

$$d^2=4x^2(2-\sqrt3).$$

This shows how a spatial problem can lead to an exact algebraic answer. The same approach works when a non-right-angled triangle provides a length needed later in a right-angled one: identify the triangle, choose a rule from its known information, then transfer the result to the next triangle.