L6 · Multi-step trigonometry problems
“I can solve problems built from two right triangles that share a side: I find the shared side first, carry its unrounded value into the second triangle, and finish — including indirect measurement with a clinometer, and two-observation problems where I set up an equation from both sightings.”
A vertical pole stands on level ground. From a point 8 m from the pole's base, on one side, a guy wire runs from the ground point to the top of the pole at an angle of 50°. On the opposite side, a second wire runs from the top to a point 11 m from the base. (a) Find the pole's height. (b) Find the angle the second wire makes with the ground. Step 1: Triangle 1 (wire 1, pole, ground): tan 50° = h/8 -> h = 8 × tan 50° ≈ 9.534 m. Keep 9.534 in the calculator — do not round yet. Step 2: Triangle 2 shares the pole: tan θ = h/11 = 9.534/11 ≈ 0.867. Step 3: θ = tan⁻¹(0.867) ≈ 41°. Height to report: 9.5 m. Say it plainly: the shared side is found once and spent twice. Rounding it in the middle taxes both answers.
Why do the two parts of this problem have to be done in this order?
A tree is sighted from 18 m away at an elevation of 42°. Find its height; later that day the sun's elevation is 65°, find the tree's shadow. -> Height: h = 18 × tan 42° ≈ 16.2 m. The tree's shadow is h ÷ tan 65° = 16.21 ÷ 2.145 ≈ 7.6 m.
"Rectangle 3.6 m by 2.4 m. Diagonal: √(3.6² + 2.4²) = ___ m. Angle between the diagonal and the long side: tan⁻¹(2.4 ÷ 3.6) = ___°." -> 4.3 m; 34°
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.