Mathematics • Year 9 • Unit 3 • Lesson 19
Multi-Step Trigonometry Problems
Build fluency with two-step problems: find an intermediate length first (using trig or Pythagoras), then use that intermediate to find the final answer. Keep full calculator precision through the middle; round only at the end.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A wire is anchored at the top of a flagpole and stretched to the ground 8 m from the base. The wire makes an angle of 55° with the ground. Find (a) the height of the flagpole and (b) the length of the wire, to 2 d.p.
Step 1 — Sketch and label.
Draw a right triangle. Flagpole = vertical leg (height h, unknown). Ground from base to wire-anchor point = horizontal leg = 8 m. Wire = hypotenuse. Angle 55° sits at the ground-anchor end of the wire.
Reason: every multi-step problem starts with a clear diagram — saves choosing the wrong ratio.
Step 2 — Tag each side relative to the 55° angle.
From the 55° angle: flagpole (height) = OPPOSITE; ground (8 m) = ADJACENT; wire = HYPOTENUSE.
Reason: opposite is across from the angle; adjacent is next to it (not the hypotenuse).
Step 3 — Find the flagpole height (opp + adj → tan).
tan 55° = h / 8 ⇒ h = 8 tan 55° ≈ 8 × 1.4281 ≈ 11.4250...
Reason: opp / adj uses tan (TOA). Keep full precision — don't round yet.
Step 4 — Find the wire length.
Method 1 (use the unrounded h): wire = √(h² + 8²) = √(11.4250² + 64) = √(130.53 + 64) ≈ 13.9486...
Method 2 (use cos directly): adj / hyp = cos 55°, so wire = 8 / cos 55° ≈ 8 / 0.5736 ≈ 13.9476...
Both methods agree to 4 d.p. — small differences come from rounding tan 55° mid-way.
Reason: two valid paths exist; Method 2 avoids carrying any rounded number forward.
Answer: flagpole height ≈ 11.43 m; wire length ≈ 13.95 m (rounded only at the end).
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A rectangular box has dimensions 6 cm by 8 cm by 10 cm. Find (a) the diagonal of the base and (b) the space diagonal (corner to opposite corner of the box), to 2 d.p.
Step 1 — Sketch: base is a rectangle 6 cm × 8 cm. Box height = 10 cm. Space diagonal goes from one bottom corner to the OPPOSITE top corner.
Step 2 — Base diagonal (Pythagoras on the base):
dbase = √( __________ ² + __________ ²) = √( __________ + __________ ) = √ __________ = __________ cm
Step 3 — Set up the space-diagonal right triangle:
Right triangle has horizontal leg = base diagonal = __________ cm; vertical leg = box height = __________ cm; hypotenuse = space diagonal.
Step 4 — Space diagonal (Pythagoras again):
dspace = √( __________ ² + __________ ²) = √( __________ + __________ ) = √ __________ ≈ __________ cm
Answer: base diagonal = __________ cm; space diagonal ≈ __________ cm.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single-step recall). The middle two are standard (two-step). The last two are extension (surveyor / 3D).
Foundation — single-step recall
3.1 A right triangle has legs 5 and 12. Find the hypotenuse. 1 mark
3.2 From 50 m away, the elevation to a tree-top is 28°. Find the tree height (2 d.p.). 1 mark
3.3 A rectangle is 9 cm by 40 cm. Find its diagonal (it's a clean integer). 1 mark
3.4 A box has dimensions 2 by 3 by 6. Find the space diagonal (it's a clean integer). 1 mark
Standard — two-step (trig then trig, or Pythagoras then trig)
3.5 A rectangle is 9 cm by 12 cm. Find (a) the diagonal and (b) the angle the diagonal makes with the longer side, to 1 d.p. 2 marks
3.6 A 4 m ladder leans against a wall at 70° to the ground. Find (a) the height the ladder reaches up the wall, and (b) the distance from the base of the wall to the foot of the ladder, to 2 d.p. 2 marks
Extension — surveyor method / 3D
3.7 From point A on flat ground the elevation to a tower top is 20°. Moving 50 m closer (to point B), the elevation becomes 35°. Find the tower height (2 d.p.). 3 marks
3.8 A 3D box is 4 cm by 5 cm by 12 cm. (a) Find the base diagonal. (b) Find the space diagonal. (c) Find the angle the space diagonal makes with the base (1 d.p.). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (box 6×8×10)
Step 2: dbase = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.
Step 3: horizontal leg = 10 cm; vertical leg = 10 cm.
Step 4: dspace = √(10² + 10²) = √(100 + 100) = √200 ≈ 14.14 cm.
Answer: base diagonal = 10 cm; space diagonal ≈ 14.14 cm.
3.1 — Legs 5 and 12
Hyp = √(25 + 144) = √169 = 13. (Classic 5-12-13 triple.)
3.2 — 50 m, elevation 28°
h = 50 tan 28° ≈ 50 × 0.5317 ≈ 26.59 m.
3.3 — Rectangle 9 by 40
d = √(81 + 1600) = √1681 = 41. (9-40-41 triple.)
3.4 — Box 2 by 3 by 6
dspace = √(4 + 9 + 36) = √49 = 7. (Clean integer.)
3.5 — Rectangle 9 by 12
(a) d = √(81 + 144) = √225 = 15 cm (3-4-5 scaled by 3 → 9-12-15).
(b) Angle with the LONGER side (12): tan θ = 9 / 12 = 0.75. θ = tan−¹(0.75) ≈ 36.9°.
3.6 — 4 m ladder at 70°
From the 70° angle (at the foot of the ladder): wall-height = opposite; ground-distance = adjacent; ladder = hypotenuse (4 m).
(a) Wall-height = 4 sin 70° ≈ 4 × 0.9397 ≈ 3.76 m up the wall.
(b) Ground-distance = 4 cos 70° ≈ 4 × 0.3420 ≈ 1.37 m from the wall.
3.7 — Surveyor tower (20° from A, 35° from B, AB = 50)
Let d = distance from B to tower base.
From B: h = d tan 35°.
From A: h = (d + 50) tan 20°.
Equate: d tan 35° = (d + 50) tan 20° ⇒ d(tan 35° − tan 20°) = 50 tan 20°.
d = 50 tan 20° / (tan 35° − tan 20°) ≈ 50 × 0.3640 / (0.7002 − 0.3640) ≈ 18.199 / 0.3362 ≈ 54.13 m.
h = 54.13 × tan 35° ≈ 54.13 × 0.7002 ≈ 37.91 m (tower height, 2 d.p.).
3.8 — Box 4 by 5 by 12
(a) Base diagonal = √(16 + 25) = √41 ≈ 6.40 cm.
(b) Space diagonal = √(16 + 25 + 144) = √185 ≈ 13.60 cm.
(c) Angle space-diagonal makes with the base: tan θ = 12 / √41, so θ = tan−¹(12 / 6.4031) ≈ tan−¹(1.8741) ≈ 61.9° (1 d.p.). Use the unrounded base diagonal to keep precision.