0580 · 6.6Pythagoras’ theorem and trigonometry in 3D

Finding triangles in 3D

0:00A straight cable stretches from one corner of a room to the opposite corner near the ceiling. Its length depends on the room’s width, length and height. A triangle drawn on the floor cannot give the whole answer. We need to connect that floor triangle to another triangle standing upright. Three dimensional problems become manageable when we find the right two dimensional triangles inside them. We will calculate lengths through solids and find the angle between a line and a plane.

0:24The important skill is choosing the correct triangle. Once that triangle is clear, familiar methods do the calculation. We need Pythagoras’ theorem and the right angled trigonometric ratios. Pythagoras says that the hypotenuse squared equals the sum of the squares of the other two sides. The hypotenuse is opposite the right angle. Sine, cosine and tangent connect an angle with ratios of side lengths. These methods still work when a triangle stands upright or slopes through a solid.

0:50Its orientation does not change its mathematical properties. However, Pythagoras and these trigonometric ratios require a right angled triangle. We must establish the right angle from the geometry, not from how the drawing looks. A plane is a flat surface extending in every direction. A rectangular face of a cuboid lies in a plane. A face diagonal joins opposite corners of one rectangle on the surface. A space diagonal joins opposite vertices through the interior of the solid.

1:18In our cuboid, the line from P to R is a face diagonal across the floor. The line from P to V is a space diagonal. The vertical edge from R to V is perpendicular to the floor plane. It therefore meets the floor diagonal at a right angle.

Cuboid space diagonal

1:32Our first exam-style example asks us to calculate the space diagonal from P to V. Its length is six centimetres, its width is eight centimetres, and its height is twenty-four centimetres. We first calculate the floor diagonal. That gives the horizontal side of the triangle containing the space diagonal. In triangle PQR, the floor diagonal squared equals 6² + 8². Evaluating the squares gives 36 + 64. Their sum is one hundred.

2:00Taking the positive square root gives a floor diagonal of ten centimetres. In triangle PRV, the space diagonal squared equals 10² + 24². 10² is one hundred, and 24² is five hundred and seventy-six. Adding them gives six hundred and seventy-six. Its positive square root is twenty-six, so the space diagonal measures twenty-six centimetres. Notice that we used two triangles, rather than inventing a new theorem. The first combines the length and width.

2:30The second combines that result with the height. Each triangle lies in a flat plane, even though the complete shape is three dimensional. For a cuboid, this gives a useful shortcut. The square of the space diagonal equals the sum of the squares of the length, width and height. The relationship works because those three dimensions are mutually perpendicular. It is not a rule for adding any three lengths shown on a solid.

Angle between a line and plane

2:52The angle between a line and a plane needs a careful definition. It is the angle between the line and its orthogonal projection onto the plane. To construct that projection, drop perpendiculars from points on the line to the plane. The resulting line shows the original line’s direction within the plane. Imagine the sloping line casting a shadow straight down onto the floor. That shadow represents its projection when the light travels perpendicular to the floor.

3:17Here, P already lies on the plane, while V projects onto R. The required angle is therefore angle VPR. This angle measures how steeply the line rises above the plane. We do not choose an arbitrary edge lying on the floor. The floor diagonal is correct because it is the projection of the particular line we are studying. The angle with the vertical measures something different. Our second example asks us to calculate the angle between the space diagonal and the base plane.

3:44This cuboid has base dimensions eight centimetres and six centimetres, with height five centimetres. The diagonal runs from P to V. Its projection onto the base is the line from P to R. The projection squared equals 8² + 6². The squares give 64 + 36. Their sum is one hundred. Taking the positive square root gives a projection length of ten centimetres. Relative to the required angle, the vertical height is opposite and the projection is adjacent.

4:11The tangent of the angle therefore equals 5 ÷ 10. This simplifies to zero point five. The inverse tangent of zero point five gives twenty-six point six degrees, to one decimal place. We did not need the space diagonal’s length because tangent uses the two sides already available. Calculating an extra length would add work without helping. Sine or cosine could also work if their required sides were known. Choose the ratio from the triangle’s information, rather than using one automatically.

Square-based pyramid example

4:40Our third exam-style example uses a square-based pyramid. Each base edge measures twelve centimetres, and each sloping edge measures ten centimetres. The apex V is directly above O, the centre of the base. Calculate the perpendicular height and the angle between a sloping edge and the base plane. The perpendicular height runs from V to O, not along a sloping edge. The projection of the edge from P to V is the line from P to O.

5:04Because O is the square’s centre, that projection is half the base diagonal. It is not half a base edge. The base diagonal squared equals 12² + 12². Evaluating gives 144 + 144. The sum is two hundred and eighty-eight. The projection is half the diagonal, so its square is the diagonal squared divided by four. Dividing by four gives seventy-two, so the projection is √72 centimetres. The sloping edge is the hypotenuse of triangle POV.

5:35Height squared therefore equals 10² − 72. Evaluating the square gives 100 − 72. Subtracting leaves twenty-eight. The height is √28, which rounds to five point two nine centimetres to three significant figures. The height is opposite the required angle, and the sloping edge is the hypotenuse. The sine of the angle therefore equals √28 ÷ 10. Applying inverse sine gives thirty-one point nine degrees, to one decimal place. We use the exact square root in this calculation, not the rounded height.

6:07A common mistake is using a sloping edge as the perpendicular height. Another is choosing a base edge instead of the line’s projection. Both mistakes create the wrong triangle. Mark the perpendicular foot first, then connect it to the point where the sloping line meets the plane. Perspective drawings can make right angles look acute or obtuse. Use the stated properties of the solid to justify them. Also check which side is the hypotenuse before using Pythagoras.

6:32Finding a shorter side requires subtraction, as the pyramid example demonstrated. On Paper Two, the non-calculator paper, you must set up these triangles and handle suitable exact calculations without a calculator. The first example uses square numbers that can be evaluated directly. On Paper Four, use degree mode for numerical trigonometry. Inverse tangent finds an angle; ordinary tangent finds a ratio. In examination working, show the projection calculation before the angle calculation.

7:01State the trigonometric relationship using the correct sides. Give lengths to three significant figures and angles to one decimal place unless instructed otherwise. Keep exact values or full calculator precision until the final answer. The central idea is that a three dimensional problem contains useful two dimensional triangles. Find a perpendicular, identify the projection, and draw the triangle containing the required length or angle. Pythagoras connects its side lengths. Trigonometry connects those lengths to its angles.

7:31For the angle between a line and a plane, always use the line’s projection onto that plane. In a cuboid, this may be a face diagonal. In a pyramid, it may run from a base vertex to the centre. Understanding that connection lets you solve the problem rather than guess a formula.