An angle of elevation starts at the observer’s eye level. If the observer and the object stand on level ground, tangent gives the vertical rise from the eyes to the top. Add eye height only when the question asks for the full height above the ground.
The 2027 SEC G3 Mathematics K310 syllabus covers elevation and depression. This guide shows how to label the triangle, choose the correct vertical reference and check your answer.
- Draw a horizontal line at the observer’s eye level.
- Measure the angle up from that horizontal.
- Use the horizontal distance as the adjacent side.
- Convert the vertical rise to the height asked for.
CHAPTER 1 OF 4 · Back to contents
Label the two heights
Draw the line of sight from the observer’s eye to the top of the object. Then draw the eye-level horizontal until it meets the object. The section of the object above that horizontal is the opposite side of the right triangle.
For a vertical object on level ground, the horizontal distance from observer to object is the adjacent side. The line of sight is the hypotenuse. If a question supplies the line-of-sight distance instead, sine relates that distance to the vertical rise.
Label the ground separately. The triangle’s vertical side stops at eye level, while the full object height runs down to its base.
CHAPTER 2 OF 4 · Back to contents
Worked example | The top of a flagpole
A student stands 18 m from a vertical flagpole on level ground. The student’s eyes are 1.60 m above the ground and the angle of elevation to the top is 34°. Find the height of the flagpole.
| Reference | Height above ground |
|---|---|
| Base of flagpole | 0 m |
| Eye-level horizontal | 1.60 m |
| Top of flagpole | 1.60 + 18 tan 34° m |
The rise above eye level is 18 tan 34° ≈ 12.141 m. Add 1.60 m to obtain a total height of approximately 13.7 m to three significant figures.
If the question asks how far above the student’s eyes the flagpole top is, stop at the vertical rise of about 12.1 m. Both results are correct for their own reference points.
Contents · Previous chapter · Next chapter
CHAPTER 3 OF 4 · Back to contents
Practise finding the distance instead
A tower is 19.0 m tall on level ground. A pupil’s eyes are 1.50 m high and the angle of elevation is 40°. The rise used in the triangle is 19.0 − 1.50 = 17.5 m.
Let d be the horizontal distance. Then tan 40° = 17.5/d, so d = 17.5/tan 40° ≈ 20.9 m. Check by multiplying the unrounded distance by tan 40° to recover 17.5 m.
If the observer stands on a raised platform, use the eye elevation relative to the base of the object. The platform height matters whenever it changes that vertical reference.
Contents · Previous chapter · Next chapter
CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Where is the angle of elevation drawn?
At the eye, between the horizontal and the upward line of sight.
Do I round the rise before adding eye height?
Keep the unrounded rise for the final addition. Round the final answer as requested.
How do I check the calculator mode?
Use degree mode for a degree angle. For a quick check, tan 45° should equal 1.
Contents · Previous chapter · Continue with the related guide