Solve Equations with Inverse Tangent

10th Grade

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Quizzes Created: 8156 | Total Attempts: 9,588,805

| Attempts: 15 | Questions: 20 | Updated: Jan 22, 2026

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1) A right triangle has opposite side 7 and adjacent side 7 relative to angle θ. Express θ using inverse tangent.

Arctan(0)

Arctan(1)

Arctan(7)

Arctan(14)

Use the tangent ratio: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 7 / 7

tan(θ) = 1

Then, to express θ, take the inverse tangent:

θ = arctan(1)

Since arctan(1) = 45°, this is the correct expression for θ.

Explanation

Submit

About This Quiz

Solve Equations With Inverse Tangent - Quiz

Ready to explore how the arctangent (arctan) function links slopes, ratios, and angles. This quiz focuses on evaluating arctan values, identifying the correct principal range (–π/2 to π/2), and interpreting what each result represents on the unit circle. You will practice determining when tangent is positive or negative, finding corresponding... see moreangles in radians or degrees, and applying tan(arctan x) = x to verify results. By the end, you will build a strong conceptual understanding of inverse tangent relationships.
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2) Evaluate arctan(−√3/3) in radians.

−π/6

−π/4

−π/3

−π/2

tan(π/6) = √3/3.

Since the value is negative, the angle is −π/6 in the range (−π/2, π/2).

Thus, arctan(−√3/3) = −π/6.

Explanation

tan(π/6) = √3/3.

Since the value is negative, the angle is −π/6 in the range (−π/2, π/2).

Thus, arctan(−√3/3) = −π/6.

Submit

3) What is the range of y = arctan(x)?

[0, π]

(−π/2, π/2)

[−π/2, π/2]

(−∞, ∞)

The arctangent function is restricted so it returns one unique angle.

That range is (−π/2, π/2), because tangent is undefined at ±π/2.

Hence, the correct range is (−π/2, π/2).

Explanation

The arctangent function is restricted so it returns one unique angle.

That range is (−π/2, π/2), because tangent is undefined at ±π/2.

Hence, the correct range is (−π/2, π/2).

Submit

4) If tan(θ) = 0.75 and θ = arctan(0.75), what is θ to the nearest tenth of a degree?

36.9°

45.0°

53.1°

60.0°

We want the angle whose tangent is 0.75.

Since tan(36°) is slightly less than 0.75 and tan(37°) is slightly greater than 0.75, the correct angle is close to 37°.

Rounded to the nearest tenth of a degree, arctan(0.75) ≈ 36.9°.

Explanation

Submit

5) Evaluate tan(arctan(−2)).

−2

−1/2

1/2

tan and arctan are inverse functions.

So tan(arctan(x)) = x for any real x.

Hence, tan(arctan(−2)) = −2.

Explanation

tan and arctan are inverse functions.

So tan(arctan(x)) = x for any real x.

Hence, tan(arctan(−2)) = −2.

Submit

6) Solve 3tan(x) − √3 = 0 for the principal value of x.

X = π/4

X = π/3

X = π/6

X = arctan(√3)

Start with the equation: 3tan(x) − √3 = 0

First, add √3 to both sides:

3tan(x) = √3

Then, divide both sides by 3:

tan(x) = √3 / 3

Then, we use the special-angle fact:

tan(π/6) = √3 / 3

So the principal-value angle satisfying tan(x) = √3/3 is:

x = π/6

This value lies in the arctan principal range (−π/2 to π/2).

Explanation

Submit

7) Evaluate arctan(1) in radians.

π/4

π/6

π/3

π/2

The question asks for the angle whose tangent is 1.

Since tan(π/4) = 1 and π/4 lies within the range (−π/2, π/2),

arctan(1) = π/4.

Explanation

The question asks for the angle whose tangent is 1.

Since tan(π/4) = 1 and π/4 lies within the range (−π/2, π/2),

arctan(1) = π/4.

Submit

8) Solve for x in the principal range: tan(x) = −√3.

−π/6

−π/4

−π/3

−π/2

We know tan(π/3) = √3.

For a negative value, reflect across the x-axis → tan(−π/3) = −√3.

Since −π/3 is within the range (−π/2, π/2), x = −π/3.

Explanation

We know tan(π/3) = √3.

For a negative value, reflect across the x-axis → tan(−π/3) = −√3.

Since −π/3 is within the range (−π/2, π/2), x = −π/3.

Submit

9) Which value is NOT in the range of y = arctan(x)?

π/2

−π/3

π/6

The arctan range is (−π/2, π/2), and the endpoints are excluded.

Thus, 0, −π/3, and π/6 are valid, but π/2 is not in the range.

Explanation

The arctan range is (−π/2, π/2), and the endpoints are excluded.

Thus, 0, −π/3, and π/6 are valid, but π/2 is not in the range.

Submit

10) Evaluate arctan(√3) in degrees.

30°

45°

60°

90°

tan(60°) = √3.

Therefore, arctan(√3) = 60°, which lies in the valid range (−90°, 90°).

Explanation

tan(60°) = √3.

Therefore, arctan(√3) = 60°, which lies in the valid range (−90°, 90°).

Submit

11) Evaluate arctan(0) in radians.

π/4

π/2

tan(0) = 0.

So the angle whose tangent equals 0 is 0.

Therefore, arctan(0) = 0.

Explanation

tan(0) = 0.

So the angle whose tangent equals 0 is 0.

Therefore, arctan(0) = 0.

Submit

12) Evaluate arctan(−1) in radians.

−π/4

−π/3

−π/6

tan(−π/4) = −1.

Since −π/4 is within (−π/2, π/2), arctan(−1) = −π/4.

Explanation

tan(−π/4) = −1.

Since −π/4 is within (−π/2, π/2), arctan(−1) = −π/4.

Submit

13) Solve for x (principal value): tan(x) = 5.

X = −arctan(5)

X = arctan(5)

X = π − arctan(5)

No solution

arctan gives the unique angle between −π/2 and π/2 whose tangent is 5.

Thus, x = arctan(5).

Explanation

arctan gives the unique angle between −π/2 and π/2 whose tangent is 5.

Thus, x = arctan(5).

Submit

14) Evaluate arctan(4/3) to the nearest 0.01 radians.

0.64

0.73

0.93

1.33

We want the angle whose tangent is 4/3 ≈ 1.33.

tan(0.79 rad) ≈ 1

tan(1.05 rad) ≈ 1.73

Since 1.33 lies between 1 and 1.73, the angle must be between 0.79 and 1.05 radians.

The closest value is about 0.93 radians.

Explanation

Submit

15) Which identity is always true for all real x?

Arctan(tan(x)) = x

Arctan(x) + arctan(1/x) = π/2 for all x

Arctan(x) = arcsin(x)

Tan(arctan(x)) = x

tan(arctan(x)) = x is always true because tan(arctan(x)) simply returns the original input x.

Whereas arctan(tan(x)) = x is not always true because arctan(tan(x)) only equals x when x is in the principal range (−π/2, π/2). Also, arctan(x) + arctan(1/x) = π/2 for all x is not always true; it holds only when x > 0.

and lastly arctan(x) = arcsin(x) is also false because arctan(x) and arcsin(x) are different functions with different domains and ranges.

Explanation

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16) Solve for θ in (−π/2, π/2): 2tan(θ) = 2√3.

θ = π/4

θ = π/3

θ = π/6

θ = arctan(2√3)

Simplify: 2tan(θ) = 2√3 → tan(θ) = √3.

The angle whose tangent equals √3 is π/3.

Hence, θ = π/3.

Explanation

Simplify: 2tan(θ) = 2√3 → tan(θ) = √3.

The angle whose tangent equals √3 is π/3.

Hence, θ = π/3.

Submit

17) If arctan(a) = π/4, find a.

1/2

√3/2

√2/2

Apply tangent to both sides: a = tan(π/4).

Since tan(π/4) = 1, a = 1.

Explanation

Apply tangent to both sides: a = tan(π/4).

Since tan(π/4) = 1, a = 1.

Submit

18) A ramp rises 18 inches over a horizontal run of 36 inches. Let θ be the angle of elevation. Which expression gives θ?

Arctan(36/18)

Arctan(2)

Arctan(1/3) + π

Arctan(18/36)

tan(θ) = rise/run = 18/36 = 1/2.

So θ = arctan(18/36).

This gives the correct angle of elevation.

Explanation

tan(θ) = rise/run = 18/36 = 1/2.

So θ = arctan(18/36).

This gives the correct angle of elevation.

Submit

19) Evaluate arctan(√3/3) in degrees.

30°

45°

60°

15°

We use a known special-angle fact: tan(30°) = √3/3.

So the angle whose tangent equals √3/3 is 30°.

This value is also within the principal range of arctan, which is between −90° and 90°.

Explanation

We use a known special-angle fact: tan(30°) = √3/3.

So the angle whose tangent equals √3/3 is 30°.

This value is also within the principal range of arctan, which is between −90° and 90°.

Submit