Chapter 12.2 — Exercise 12.1 — Single Triangle Problems

A Tower, Directly

A tower stands vertically, 15 m from an observation point, with the top's angle of elevation measured at exactly 45°.

A 45° angle of elevation always means the height and the horizontal distance are exactly equal, since tan45°=1 — worth recognising on sight the next time a problem hands over a 45° angle, without needing to work through the substitution at all. No other angle in the standard specific-angle table produces this same "height equals distance" shortcut, which is exactly why 45° problems tend to resolve noticeably faster than any other angle appearing anywhere in this whole exercise.

A Broken Tree, Two Segments Added

A tree breaks during a storm and its top touches the ground 6 m from its own base, making a 30° angle with the ground. The original full height is the broken segment's length plus the still-standing stump's height — two entirely separate calculations from the same triangle, added together at the very end.

It's worth being clear about what each of these two quantities actually represents, since they're easy to conflate: tan30° finds the height of the point where the tree snapped (the stump that's still standing), while cos30° finds the full length of the broken, now-slanted piece lying against the ground. Adding a height to a slant length only works here because that slanted piece, stood back upright, would reach exactly as high as it currently measures along its own length.

It's also worth noticing which side plays which role changes between the two calculations, even though both come from the same triangle and the same 30° angle. The first uses the stump as the side opposite the angle and the 6 m ground distance as the adjacent side, giving tangent. The second uses that same 6 m ground distance, but now as the side adjacent to the angle relative to the hypotenuse — the broken piece itself — giving cosine instead. One triangle, one angle, two different pairs of sides, two different ratios.

Six More Single-Triangle Problems

The rope problem is worth a second look for its angle placement: the 30° sits at the top of the pole, between the pole itself and each rope, which makes the pole the side adjacent to that angle rather than the side opposite it — exactly why cosine, not tangent, is the ratio that connects them. Reading where an angle actually sits in a described scenario — at the top, at the base, at an observer's eye — matters just as much as reading which lengths are given.

The final area problem is a genuinely different kind of question from the other five: instead of finding a length or an angle, it uses a trigonometric ratio purely as a stepping stone toward the ordinary ½×base×height area formula from earlier coursework. Whenever a triangle's area needs finding from two sides and the angle between them, dropping a perpendicular and using sine to find that perpendicular's length is the standard route in — a technique worth remembering as its own reusable shortcut, not just something specific to this one problem.

A Ladder, With a Correction Worth Flagging

An electrician needs to reach a point 1.8 m below the very top of a 9 m pole, climbing a ladder set at 60° to the ground. The repair point itself sits 9−1.8=7.2 m above the ground.

The ladder length is worth stating precisely as 4.8√3 m rather than a bare decimal — written as a plain number without its root symbol, "4.8√3" can look deceptively like "4.83," a genuinely different and smaller value. Keeping a surd's root symbol attached all the way to the final answer avoids exactly this kind of misreading, and multiplying it out to roughly 8.31 m at the very end confirms the two aren't remotely the same length.

An Observer's Own Height, Included

An observer who stands 1.8 m tall is positioned 13.2 m from a palm tree, measuring the top's angle of elevation at 45° from eye level — not from the ground below.

This is the one problem in the exercise where the third rule from the chapter introduction — observer height ignored unless stated — genuinely comes into play. The triangle itself only ever measures the tree's height above eye level; the observer's own 1.8 m has to be added back on afterward as a separate, non-trigonometric step, easy to forget entirely if the final answer is taken straight from the triangle alone.

It's worth noticing, too, that the observer's height and the tree's height genuinely stay separate calculations throughout — 1.8 m never enters the triangle itself, since it sits below eye level and outside the angle of elevation entirely. Only after tan45° has already produced the 13.2 m figure does 1.8 m get reintroduced, and only by plain addition at the very end, not through any further trigonometry or ratio work whatsoever.

From One Triangle to Two

Every single scenario in this exercise fit inside just one right triangle, however differently each one was dressed up on the page. Exercise 12.2 moves to problems where one triangle genuinely isn't enough information — two separate angles, observed from two separate positions, needing two triangles solved together. For the elevation/depression definitions and the three problem-solving conventions this entire exercise leaned on throughout, revisit the chapter introduction once more before starting the next one.