SME
Grade 12 Math · Problem type
Inverse tangent: the angle of a slope
Use this for “A road rises 3 m over 40 m of ground. What angle does it make with the level?”
Chart
Horizontal asymptotes y = −90 and y = 90
Range: (−90°, 90°)
f(0.075) ≈ 4.2892°
This is an example. Type your own values to start a new problem.
tan⁻¹
rise (m)
run (m)
=
A
°
x
Rise over run
worked out (not typed)
Formulas
x
=
rise
run
0.075
=
3
40
A
=
tan⁻¹(
x
)
4.2892 = tan⁻¹(0.075)
Assumptions
•
The angle of a slope is tan⁻¹ of rise over run.
•
tan⁻¹(x) is the one angle between −90° and 90° whose tangent is x; every x has one.
•
The curve nears ±90° but never reaches them: those are asymptotes.
Step-by-step
WE KNOW
rise = 3 m, run = 40 m
FIND
rise over run (x), angle of the slope (A)
Step 1 · Find rise over run (x)
Use x = rise ÷ run
The slope is the rise for each 1 of run.
x
=
3
40
x
=
0.075
Step 2 · Find angle of the slope (A)
Use A = tan⁻¹(x)
The angle between −90° and 90° whose tangent is rise over run.
A
=
tan⁻¹(0.075)
A
=
4.2892°
Check
Put the numbers back in. Both sides should match.
0.075
=
3
40
✓
tan(4.2892°) = 0.075 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.