Every 3D problem is a 2D problem in disguise. Find the right-angled (or non-right-angled) triangle hiding inside the solid, redraw it flat on paper with its vertices labelled, then solve it using ordinary 2D tools.

Use Pythagoras twice. First find the diagonal of the base rectangle: $d_{\text{face}}^2=l^2+w^2$. Then treat that face diagonal as one leg of a new right-angled triangle standing up inside the cuboid, with the cuboid's height as the other leg:

1. Find the point of contact — where the line meets the plane.
2. Pick the other end of the line.
3. Project that point straight down onto the plane (perpendicular drop).

These three points form a right-angled triangle (right-angled at the projected point), and the angle you want is at the point of contact.

Worked example: Find the angle the space diagonal $AG$ makes with base $ABCD$ in a cuboid.

• Point of contact: $A$. Other end: $G$. Projection of $G$ onto the base: $C$.
• Triangle $AGC$ is right-angled at $C$ (since $CG$ is vertical, perpendicular to the whole base).
• $\tan\theta=\dfrac{CG}{AC}$ — but $AC$ usually isn't given directly, so find it first from the base rectangle via Pythagoras, then substitute.

A triangle "floating" between three vertices of a solid (e.g. connecting points on different faces) is often scalene, with no right angle at all. SOHCAHTOA and Pythagoras do not apply here — you need the sine or cosine rule, exactly as in 2D:

• Know all 3 sides (SSS) → cosine rule, to find an angle.
• Know 2 sides + included angle (SAS) → cosine rule, to find the third side.
• Have a complete opposite pair → sine rule.

A square-based pyramid has base side $8$ cm and all four slanted edges $10$ cm. Find the angle between a slanted edge and the base.

1. This is a line-and-plane problem: line $AX$ (slanted edge), plane = base. Point of contact $A$; projection of apex $X$ is the base's centre $M$. Triangle $AMX$ is right-angled at $M$.
2. Need $AM$ first (half the base diagonal). Base diagonal: $AC=\sqrt{8^2+8^2}=\sqrt{128}$, so $AM=\dfrac{\sqrt{128}}{2}\approx5.657$ cm.
3. In $\triangle AMX$: hypotenuse $AX=10$, adjacent $AM\approx5.657$.

Work backwards from the target: identify the triangle containing what you want, check if all its sides/angles are known — if not, that missing piece is itself the answer to a smaller sub-problem in a different triangle. Solve that one first, then return.

• Applying SOHCAHTOA or Pythagoras to a triangle that only looks right-angled in a 3D sketch.
• Measuring the angle to the wrong point instead of the true perpendicular projection.
• Forgetting to find a missing side via Pythagoras before using it in the cosine or sine rule.
• Calculator left in radian mode.