Finding lengths in 3D using Pythagoras and trigonometry

A mast stands in one corner of a rectangular field, forty metres by thirty metres. From the corner diagonally opposite, you look up at the top of the mast at an angle of eighteen degrees. That angle is measured along the field's diagonal, and nobody has told you how long the diagonal is. How tall is the mast? It takes a chain of two triangles. Pythagoras finds one length, and trigonometry uses that length to find the next.

This is video nine of eleven in Pythagoras, Trigonometry and Coordinate Geometry. Spotting the flat triangle inside a solid comes from Pythagoras' theorem in three dimensions, so go back to that one if it's new.

Chains like this are Higher-tier work, or Extended on Core and Extended courses, and specifications place them differently, so look at your own.

A three-D length question is usually a chain. One flat triangle gives you a length, and that length becomes a side in the next triangle. The chapter's signal picks the tool for each triangle. An angle given or wanted means trigonometry; no angle anywhere means Pythagoras. One triangle stands upright: the mast, the line of sight from the far corner up to its top, and the field's diagonal along the ground. The eighteen-degree angle of elevation sits at the far corner. The other, meanwhile, lies flat on the field: the forty-metre side, the thirty-metre side, and the diagonal joining their ends. Which triangle comes first: the upright one, or the one lying on the field? The flat one. The upright triangle has an angle but no known length, and trigonometry can't start without one. The flat triangle has two known sides, so Pythagoras finishes it.

In the flat triangle, the square corner is where the forty and the thirty meet, so the diagonal is the longest side. Add the squares: forty squared is one thousand six hundred, and thirty squared is nine hundred. One thousand six hundred plus nine hundred is two thousand five hundred. The square root of two thousand five hundred is fifty, so the diagonal is exactly fifty metres. Every angle faces its own side - look straight across. In the flat triangle, stand in the square corner and you see the diagonal, so there it's the hypotenuse, the longest side. In the upright triangle, though, the square corner is at the mast's foot, and from there you see the line of sight instead. So the same fifty metres is the hypotenuse in one triangle - and a shorter side in the next. Label each triangle on its own, from scratch.

Right, the upright triangle comes next. Label it from the eighteen degrees at the far corner. The line of sight faces the square corner, so it's the hypotenuse. The mast's height faces the angle, so it's opposite. The fifty-metre diagonal runs from the angle to the square corner, so it's adjacent. Opposite wanted, adjacent known. Which ratio: sine, cos or tan? It's tan, the one ratio that uses opposite with adjacent. Tan eighteen equals the height over fifty. Multiply both sides by fifty, and the height equals fifty times tan eighteen. Press fifty, times, tan, eighteen, equals. The display reads sixteen point two four five nine eight four eight one. To one decimal place, the mast is sixteen point two metres tall. A student writes: tan eighteen equals the height over forty. What has that student got wrong in their tan line? It mixes triangles. That side of the field lies flat on the ground, but the angle belongs to the upright triangle, where the side next to it is the fifty-metre diagonal. Draw the two triangles separately, side by side. A length crosses from one to the other only once it has been found, like the fifty metres did.

Rounding too early is the other way a chain breaks, and a pyramid shows it. A square-based pyramid has a base of side ten metres. Each sloping edge makes an angle of sixty-two degrees with the base. That angle sits at a base corner, between the edge and half the base diagonal, the line on the base running from that corner to the centre. Why half the diagonal is the right line for that angle is the next video's topic. Here the angle is given, and a length is wanted: the sloping edge. Which comes first in this chain: Pythagoras or trigonometry? Pythagoras, on the base, because the upright triangle has no side known yet. The diagonal is the longest side of a flat triangle with two ten-metre sides. Ten squared plus ten squared is a hundred plus a hundred, which is two hundred. The diagonal is root two hundred. Press the square root key, two hundred, equals, and the display shows ten root two. The S to D key turns it into fourteen point one four two one three five six two. Divide that by two for half the diagonal: press divide, two, equals, then the S to D key. The display reads seven point zero seven one zero six seven eight one two metres, and every digit stays. Now the upright triangle. Stand in the sixty-two degrees at the base corner. Half the diagonal runs from the angle to the square corner under the apex, so it's adjacent, and the sloping edge is the hypotenuse. Adjacent known, hypotenuse wanted. Which ratio now? Cos this time, because cos pairs adjacent with hypotenuse. Cos sixty-two equals half the diagonal over the edge. The edge is on the bottom, so multiply both sides by the edge, then divide both sides by cos sixty-two. That leaves the edge equals half the diagonal divided by cos sixty-two. Press the answer key for the full half-diagonal, divide, cos, sixty-two, equals. The display reads fifteen point zero six one seven five nine five nine. To one decimal place, each sloping edge is fifteen point one metres. Now, round too early and see what happens. Take the diagonal as fourteen point one, halve it to seven point zero five, and divide by cos sixty-two. The display reads fifteen point zero one six eight eight four, which rounds to fifteen point zero. Why did one early rounding change the final answer? The true edge sits only just above the halfway mark between two answers. Losing a few hundredths on the way pushes it below that mark, so it rounds down instead of up. One examiner's report describes responses that started with Pythagoras and then "lost accuracy by premature rounding." Premature means too soon. The fix is to keep the full display, or the exact root, all the way along the chain, and round once, at the very end.

The mast needed two triangles, taken in the right order, with nothing rounded in between. Now for some chains you haven't met. In a chain of two triangles, how do you decide which to solve first? Start with the one that already has enough: two sides for Pythagoras, or a side and an angle for trigonometry. Try a field twenty by fifteen metres, its mast's top seen at thirty degrees. How tall? Fourteen point four metres. The diagonal is twenty-five, and twenty-five times tan thirty is fourteen point four three. Another pyramid: base eight metres, each edge at fifty degrees to the base. How high is it? Roughly six point seven metres. Centre to corner is root thirty-two, and multiplying that by tan fifty gives the height.

If you've got this down, a thumbs-up ticks it off, so this video stays out of your way next time you revise. If the chain still slips, save it for later, then explain the mast to someone else out loud. Teaching the order to another person is what makes it stick.

Next in the chapter: Finding the angle between a line and a plane.

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