L5 · Angles of elevation and depression
“I can model a sighting problem as a right triangle: the angle of elevation is measured up from the horizontal, the angle of depression is measured down from the horizontal, and the two are equal (alternate interior angles) when the same sight line is viewed from its two ends. I can find heights and distances from them, adding eye height when the problem gives it.”
From a point 40 m from the base of a tower, the angle of elevation to the top is 37°. Find the tower's height. Step 1: Model. Ground distance = adjacent (40). Height = opposite. The line of sight is the hypotenuse. Angle at the observer: 37°, measured from the horizontal ground line. Step 2: Opposite + adjacent -> tan. tan 37° = h/40. Step 3: h = 40 × tan 37° ≈ 40 × 0.754 = 30.1 m. (Note how the numbers echo L3/L4: tan⁻¹(30/40) is how you would find the angle if the height were known instead.)
Why is tan the default first choice for elevation problems?
From 25 m away, the angle of elevation to the top of a building is 48°. Find the height. -> Height = 25 × tan 48° ≈ 25 × 1.111 = 27.8 m.
"Distance 100 m, elevation 30°: h = 100 × tan 30° = 100 × 0.577 = ___ m" -> 57.7 m
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.