Trigonometry & Heights / Distances • Topic 4 of 4
Heights & Distances
Model the situation as a right triangle: the angle of elevation (looking up) or depression (looking down) sits at the observer, the height is the opposite side and the ground distance the adjacent side, so tan(angle) = height/distance. Standard angles 30, 45, 60 give clean answers. For two observations, set up two equations and subtract.
✅ Solved examples
1. The angle of elevation of a tower top from 30 m away is 45 degrees. Tower height?
tan 45 = h/30 = 1 -> h = 30 m.
2. From 20 m, the elevation of a pole top is 60 degrees. Height?
tan 60 = h/20 = root3 -> h = 20 root3 m.
3. A 15 m ladder leans at 60 degrees to the ground. Height it reaches?
sin 60 = h/15 -> h = 15 x root3/2 = 7.5 root3 m.
4. Elevation of a tower top is 30 degrees from 90 m. Height (use tan 30 = 1/root3)?
h = 90/root3 = 30 root3 m.
✏️ Practice — try these, take hints as needed
1. Elevation 45 at 50 m. Height?
tan 45 = h/50.
—
—
50 m
2. Elevation 60 at 10 m. Height?
tan 60 = h/10.
10 root3.
—
10 root3 m
3. A 10 m ladder at 30 degrees reaches what height?
sin 30 = h/10.
10 x 1/2.
—
5 m
4. Elevation 30 at 60 m, height?
tan 30 = h/60.
60/root3.
—
20 root3 m
5. Shadow of a pole equals its height. Sun elevation?
tan = h/h = 1.
—
—
45 degrees
📝 Topic test — 8 questions
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Formula Reference Sheet
This chapter
Ratios & identities
| Basic ratios | sin = opp/hyp, cos = adj/hyp, tan = opp/adj |
|---|---|
| Reciprocals | cosec = 1/sin, sec = 1/cos, cot = 1/tan |
| Pythagorean | sin^2 + cos^2 = 1 |
| Secant | 1 + tan^2 = sec^2 |
| Cosecant | 1 + cot^2 = cosec^2 |
Standard angles & H&D
| sin | 0, 1/2, 1/root2, root3/2, 1 (at 0,30,45,60,90) |
|---|---|
| cos | 1, root3/2, 1/root2, 1/2, 0 |
| tan | 0, 1/root3, 1, root3, undefined |
| Heights & distances | tan(angle of elevation) = height / horizontal distance |
SSC reference
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