Simple 2-D problems

A right-angled triangle has exactly three named sides once you fix a reference angle $\theta$: the hypotenuse (always opposite the right angle), the opposite side (facing $\theta$), and the adjacent side (next to $\theta$, not the hypotenuse). All of trigonometry for this topic sits inside three ratios built from those three sides.

where: $\theta$ is the known (or wanted) angle, and opposite, adjacent, hypotenuse are the triangle's three sides measured in the same unit. The whole method for a simple 2-D problem is four steps done in order: draw and label the triangle from the story, mark which side is opposite/adjacent/hypotenuse relative to $\theta$, choose whichever of the three ratios above already contains the known side and the unknown quantity, then solve. When two sides are known and the third is wanted, the Pythagoras theorem, $\text{hyp}^2 = \text{opp}^2 + \text{adj}^2$, is often faster than any ratio. A quick sorting rule saves time in a hurry: if the given and wanted sides are the hypotenuse plus one leg, use sine or cosine (whichever leg you have); if both given and wanted sides are legs — no hypotenuse involved at all — use tangent. The same skeleton triangle keeps reappearing under different costumes. A ladder against a wall, a ramp climbing to a doorway, a flagpole's guy wire pegged into the ground, a kite string pulled taut — in every case one side is a slanting length (the hypotenuse), one side is vertical, and one side is horizontal, meeting at a right angle. Reading the story only has to answer two questions: which of the three lengths is given, and which one is wanted. Everything after that is mechanical — pick the ratio whose definition already contains both of those sides, substitute, and solve.

The goal is to show that, once one angle and one side of a right triangle are known, both remaining sides can be written directly in terms of that angle and side, so that "which ratio do I use" becomes a matter of reading the formula off rather than guessing.

Step 1. Fix the reference angle $\theta$ and the hypotenuse $h$ as the known quantities — this is the commonest setup, since a ladder, string, or ramp length is usually the given straight-line measurement. By the definitions above, $\sin\theta = \text{opposite}/h$ and $\cos\theta = \text{adjacent}/h$.

Step 2. Rearrange each definition for the side it contains, since $h$ and $\theta$ are known numbers. This gives $\text{opposite} = h\sin\theta$ and $\text{adjacent} = h\cos\theta$ directly — no further algebra needed once $\sin\theta$ and $\cos\theta$ are read from the standard-angle table.

Step 3. Check the result with the Pythagoras theorem, which must hold for any right triangle: $(\text{opposite})^2 + (\text{adjacent})^2 = (h\sin\theta)^2 + (h\cos\theta)^2 = h^2(\sin^2\theta + \cos^2\theta)$. Since $\sin^2\theta + \cos^2\theta = 1$ for any angle, this simplifies to exactly $h^2$, matching the hypotenuse squared — confirming the two side formulas are consistent with each other.

Step 4. Read the payoff: whenever the hypotenuse and the angle are known, both other sides follow at once from $\text{opposite} = h\sin\theta$ and $\text{adjacent} = h\cos\theta$, without needing to solve two separate equations from scratch each time. The same idea runs in reverse too: if instead the hypotenuse is unknown but one leg and the angle are given, the definitions rearrange to $h = \dfrac{\text{opposite}}{\sin\theta}$ or $h = \dfrac{\text{adjacent}}{\cos\theta}$ — the identical three-way relationship, just solved for a different letter depending on what the question happens to give you.

A ladder $10$ m long leans against a vertical wall with its foot $5$ m from the wall. Find the angle it makes with the ground and the height it reaches.

Step 1. Draw the triangle: the ladder ($10$ m) is the hypotenuse, the $5$ m foot-distance is the side adjacent to the ground angle $\theta$, and the height on the wall is the side opposite $\theta$.

Step 2. Find $\theta$ using cosine: $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{10} = \dfrac12$. Since $\cos 60^\circ = \tfrac12$, $\theta = 60^\circ$.

Step 3. Find the height using the Pythagoras theorem: $\text{height}^2 = 10^2 - 5^2 = 75$, so $\text{height} = \sqrt{75} = 5\sqrt3 \approx 8.66$ m. The ladder makes a $60^\circ$ angle with the ground and reaches $5\sqrt3 \approx 8.66$ m up the wall.

Competitive stretch

A kite's string, $80$ m long, is taut at $30^\circ$ to the horizontal ground, reaching a height $h = 80\sin30^\circ = 40$ m (the Pass-1 result). How far along the ground, horizontally, is the kite from the person holding the string?

Step 1. The horizontal distance is the side adjacent to the $30^\circ$ angle, with the string ($80$ m) as hypotenuse and the height ($40$ m) as the opposite side already found.

Step 2. Apply the Pythagoras theorem instead of a fresh ratio, since two of the three sides are now known: $\text{distance}^2 = 80^2 - 40^2 = 6400 - 1600 = 4800$.

Step 3. Solve: $\text{distance} = \sqrt{4800} = 40\sqrt3 \approx 69.3$ m. The kite is about $69.3$ m horizontally from the person holding the string. Notice what changed between the two examples: the ladder problem started with an angle unknown and used a ratio to find it, then switched to the Pythagoras theorem for the leftover side; the kite stretch started with the height already known from a ratio and used the Pythagoras theorem for the leftover side too. Once one ratio has supplied either the angle or a side, the fastest route to whatever remains is very often the Pythagoras theorem rather than a second trigonometric calculation.

• Skipping the labelled diagram — without one, "opposite" and "adjacent" are easily swapped, especially when the triangle is drawn tilted in the question's picture.
• Choosing tangent when the hypotenuse is one of the given lengths; hypotenuse problems always need sine or cosine, never tangent.
• Forgetting to rearrange before substituting — writing $\sin30^\circ = h \times 80$ instead of $\sin30^\circ = h/80$.
• Mixing units, such as leaving one side in metres and another in centimetres before applying a ratio or the Pythagoras theorem.

Ask the tutor: Stuck on a ramp, flagpole guy-wire, or ladder problem of your own — ask the AI tutor to work through the triangle with you step by step.