Two angles in one figure, and the pair of equations they hand you

0:00Here is a tower on the far bank of a canal, and here is you, on this side. You can measure the angle up to its top. You cannot measure anything else. You cannot walk to the foot of the tower, so you do not know how far away it is. One angle gives you one equation, and that equation has two unknowns in it: the height, and the distance you could not pace.

0:26One equation, two unknowns. There is nothing to solve. So you take one step backwards, and look again. That second look changes everything, and this video is about exactly what it buys you. Here is the thing to be clear about from the start. Stepping back and sighting again does not measure the distance to the tower. You still have not been to the far bank. What it gives you is a second equation, in the same two unknowns.

0:58The first station says: the height equals the near distance times the tangent of the first angle. The second says: the height equals the near distance plus your step, times the tangent of the second angle. Two equations, two unknowns, and one of those unknowns is a length nobody ever wanted. Eliminate it between the two, and the one you do want survives. That is the whole method. Every problem in this video is that same move wearing a different costume.

1:34And notice what it needs: the two angles must be different. Sight the same target twice at the same angle and the second equation is a copy of the first. Hand that to a solver and it should refuse, whichever unknown you ask it for. Two identical sightings are one sighting. The purest case is a shadow. A tower stands on level ground. When the sun is high, at sixty degrees above the horizon, the tower casts a short shadow.

2:09Later, with the sun at thirty degrees, the shadow is longer, and it is longer by forty metres. Find the height of the tower. Look at what you were given. Not the height. Not the distance to anything. One number, forty metres, and it is not a shadow. It is the gap between two shadow tips. This is the single commonest error on this problem, so let us make it concrete rather than just warn about it.

2:39Read the forty as the long shadow itself and you get a tower of forty over root three. Read it correctly, as a difference, and you get something else entirely. The two answers are different numbers, and only one of them is the tower. Call the height h and the short shadow x. The sixty degree triangle gives: h over x is root three. The thirty degree triangle has the same tower, and a base of x plus forty. It gives: h over x plus forty is one over root three.

3:17Now substitute. From the first, h is x root three. Put that into the second and multiply out: three x equals x plus forty. So x is twenty. Stop there and look at what just happened, because this is the part that surprises people. The height was what you wanted, and the height is what you substituted away. The equation you actually solved was for x, the short shadow, which nobody asked about and nobody measured.

3:51It came out at twenty metres, and the height follows from it: twenty root three. The unwanted length is not a nuisance. It is the road. It is worth seeing how badly one sighting fails, rather than being told. Take the sixty degree reading alone and ask which towers agree with it. Here are twelve of them: a short one close by, a taller one further off, and ten more, twelve different heights, and every single one casts a shadow that reads sixty degrees.

4:27One sighting does not narrow the field at all. It only fixes the ratio of height to distance. Now bring in the second sighting: the shadow forty metres longer, reading thirty. Eleven of the twelve fail it. Exactly one survives, the tower of twenty root three standing over a shadow of twenty metres. That is what the second equation does. It is not extra confidence. It is the difference between a family of answers and one answer.

5:02Now the same idea with the two angles at one eye instead of two. You stand on the ground looking at a building ten metres tall, with a flag on a pole above it. The top of the building sits at thirty degrees. The top of the flag sits at forty five. How long is the flag pole, and how far away are you? The lower triangle is ordinary: the building is ten, the angle is thirty, so your distance is ten root three.

5:33Now use that same distance in the upper sighting. At forty five degrees the height above your eye equals the distance across, so the top of the flag is ten root three up. The building takes ten of that. The flag pole is ten root three less ten. Look at the shape of that answer. The given was a height, and the answer is a difference of two heights. In the shadow problem, the given was a difference of two distances. Same machinery, mirrored.

6:07Before going further, here is a check that costs nothing and catches transposed labels. From one station, which of two targets carries the larger angle? The reflex answer is: the higher one. That answer is a trap, and here is where it breaks. State it properly: the target lying further from your own horizontal carries the larger angle. Looking up, further from your horizontal does mean higher. Looking down, it means lower.

6:40Take one station, several eye heights, several distances, and a spread of targets above and below. Compare them in pairs: over three hundred and thirty six pairs, the distance from your horizontal calls it correctly every single time, and the reverse rule is wrong every single time. The naive version does no better than a coin over the same pairs. It is right on every one of the seventy two pairs where both targets are above you, and wrong on every one of the seventy two where both are below.

7:18One rule, two answers. Learn the one that survives both. And one more thing the same test settles: at a fixed height difference, the further target is always the one seen at the smaller angle. Walking towards something raises the angle to its top. So look down instead. You are on the roof of a tall building, looking at a shorter one eight metres high. Its roof reads thirty degrees below your horizontal. Its base reads forty five.

7:52This is the figure the naive rule fails on. The roof of the short building is the higher target, and it carries the smaller angle, because it is nearer your level. Now solve it. Drop a plumb line down your own building to the level of the short roof. That splits your height into two pieces: the part above the short roof, and the eight metres alongside it. The thirty degree sighting sees the upper part across the gap. The forty five degree sighting sees the whole height across the same gap.

8:29Eliminate the gap between them, and your building comes out at four times three plus root three. And the gap between the two buildings is that same number. The building is exactly as tall as the two are far apart. That solution leaned on something worth making explicit. In the shadow problem the two triangles literally shared a side: the tower stood in both of them. Here they share nothing at all. The upper sightline runs across at roof level; the lower one runs down to the ground.

9:07Of the five figures in this video, three hold one segment in common, and two hold none. So what ties these two triangles together? Not a shared segment, but a pair of equal ones. The two horizontals are the opposite sides of a rectangle, and opposite sides of a rectangle are equal. That is checked here, not assumed: all four corners are right angles and both pairs of opposite sides match.

9:38Push one corner sideways and it fails on both counts. Shear the whole side over instead, and the opposite sides still match while not one angle is right. Two conditions, and the figure needs them both. That answer arrived through a denominator of root three less one, and it is worth watching that denominator go. Multiply above and below by root three plus one. The bottom becomes root three less one, times root three plus one.

10:12That is a difference of squares: three, less one. Two, exactly. The surd is gone. Now try the mistake. Multiply above and below by root three less one, the same thing again. The bottom is four less two root three. Still a surd. Nothing was cleared. It has to be the conjugate, the same two terms with the sign between them flipped, and only that. Do it right and eight over root three less one becomes four times root three plus one: four, plus four root three.

10:49Rationalising is not a presentation rule you follow to please someone. It is what turns an opaque quotient into a height you can read as a sum of two parts. Here is one that looks like the others and is not. A bridge deck sits three metres above both banks. From a point on the deck, one bank is at thirty degrees below level and the other at forty five. How wide is the river?

11:19Two angles, two triangles, two right angles. It looks like everything else in this video. But look at what the two triangles share: the three metre drop from the deck, straight down. And that length is given. So there is nothing to eliminate. Each triangle can be solved on its own. Three metres at thirty degrees puts one bank three root three away. Three metres at forty five puts the other bank three away.

11:50The width is the sum: three plus three root three. This is the easiest figure of the lot and it is usually taken for the hardest, purely because it has two angles in it. Two angles are not the difficulty. An unmeasured shared length is the difficulty. Once you see the skeleton, a whole family of these collapses into one problem. A canal, with a tower on the far bank. Sixty degrees from the near edge, thirty from twenty metres further back. The canal is ten metres wide and the tower is ten root three.

12:31A road eighty metres across with equal poles facing each other. From a point between them, sixty degrees one way and thirty the other. The poles are twenty root three, standing twenty metres and sixty metres from where you are. A boy walking towards a thirty metre building, the elevation climbing from thirty degrees to sixty: he walks nineteen root three. Two ships in line from a seventy five metre lighthouse, at thirty and forty five degrees: seventy five root three, less seventy five, apart.

13:09A balloon drifting level, dropping from sixty degrees to thirty across the sky: fifty eight root three. Different objects, different numbers, one move. Two equations in a shared unknown, and the shared unknown eliminated. One last problem, and it is the strongest statement of the whole idea. A car comes along a straight road towards a tower at steady speed. From the top of the tower it is seen at thirty degrees. Six seconds later, at sixty.

13:45How much longer until it reaches the foot? You are not told the height of the tower. You are not told the speed of the car. You are not told a single distance. Call the height h. The far position is h root three away, the near one is h over root three. The six seconds covered the difference between them, and the remaining distance is h over root three.

14:12The difference is exactly twice what is left, so the remaining time is half of six. Three seconds. And h walked out. Every h in the ratio cancelled. Run it on twelve different towers and the two distances are twelve different pairs of numbers, every time, while the answer is three seconds, every time. That is not a coincidence of the numbers. It is what elimination is for. You never needed the tower. You needed the relation between two sightings, and a second look at the same thing gave you it.