A map is a plan view: every distance on it is horizontal. The ground is not, and two right triangles reconcile the two.

Along the hill. $g = \sqrt{m^{2} + r^{2}}$ — g equals root, m squared plus r squared. $m$ is the map distance, horizontal, as the sheet shows it; $r$ is the rise, the height gained between the contours; and $g$ is the ground distance, the hypotenuse your boots actually walk. All three in metres. The ground distance is always the largest of the three — a slope can only ever stretch a walk, never shorten it.

Here is the number that surprises everyone. The cosine is forgiving: a 20 % grade adds only about 2 % to the distance, and it takes roughly a 30 % slope before the stretch is worth redrawing a route card for. Your legs feel every metre of the rise; the map length barely shows it. That is precisely why Naismith bills the climb in TIME instead of distance, which is the next lesson.

Up the tree. $H = d\tan\theta + e$ — H equals d tan theta, plus e. $d$ is the horizontal distance from you to the base, paced on level ground; $\theta$ — theta — is the elevation angle to the top, read off a clinometer, above horizontal; $e$ is your own eye height, ground to eye, about 1.5 m; and $H$ is the height of the object from the ground to its top. Tangent is opposite over adjacent, so $d\tan\theta$ is the height above your EYE — and the $+\,e$ puts your own body back into the answer.

Two field cautions. Pace the standoff on level ground: the tangent wants the horizontal leg, and a distance paced up a slope hands it the hypotenuse instead, which quietly inflates the height. And there is a lovely trick hiding in the formula — $\tan 45^{\circ} = 1$, so walk back and forth until the clinometer reads exactly 45°, then simply pace your distance to the trunk and add your eye height. The tree is as tall as you are far away. No trigonometry required at all.