This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! For the ramp, the angle θ satisfies tan(θ) = rise/run = 4/3, so θ = arctan(4/3) ≈ 53.1°, using the inverse tangent since we have opposite over adjacent. Choice B correctly identifies arctan for the tangent ratio and computes the angle accurately within its range. A distractor like choice D swaps the ratio to $tan^{-1}$(3/4) ≈36.9°, which would be for the wrong sides; always ensure opposite and adjacent are correctly assigned. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Great job applying this to real-world scenarios like ramps—you've got this!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! Composing sin with arcsin undoes the operation, so sin(arcsin(-0.3)) = -0.3, as -0.3 is in the domain [-1,1] and arcsin returns an angle in [-90°,90°] whose sine is -0.3. Choice B correctly evaluates the composition, recognizing that inverse functions cancel each other out. Choice D fails because inverse trig functions do accept negative inputs within their domains, allowing for angles in quadrant IV where sine is negative. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! The domain of sin(θ) is restricted for arcsin(x) because sine is not one-to-one over all reals, leading to multiple angles for one output, so restriction ensures a single-valued inverse. Choice B correctly explains that without restriction, the inverse wouldn't be a function due to multiple angles per output. Choice D fails because arcsin's domain is actually [-1,1], not all reals, as inputs must be valid sine values. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! With opposite side 4 and adjacent side 3, tan(θ) = 4/3 ≈ 1.333, so θ = arctan(4/3) ≈ 53.1°, fitting the arctan range (-90°, 90°). Choice A correctly uses arctan for the opposite-over-adjacent ratio, yielding the principal angle. Choice B fails because arcsin(4/3) has an argument >1, outside the [-1,1] domain, making it undefined. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! Evaluating arcsin(1/2) gives 30°, as sin(30°) = 1/2 and 30° is within [-90°, 90°], the principal range. Choice B correctly identifies the principal value, adhering to the range restriction for arcsin. Choice A fails because 150° is outside the arcsin range, even though sin(150°) = 1/2, but arcsin returns only the value in [-90°, 90°]. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! The notation $sin^{-1}$(x) denotes the inverse sine function, arcsin(x), which finds the angle whose sine is x, not the reciprocal. Choice B correctly interprets $sin^{-1}$(x) as the inverse function, distinguishing it from the reciprocal csc(x) = 1/sin(x). Choice A fails by confusing the inverse notation with the reciprocal, a common mistake since -1 can mean either, but in trig context, $sin^{-1}$ means inverse. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! The range of arctan(x) is -90° < y < 90° because tangent approaches ±∞ as angles approach ±90° but never reach them, ensuring a unique output for every real input. Choice C correctly states this open interval, reflecting that arctan never actually hits ±90°. A distractor like choice B uses closed intervals, but arctan doesn't include the endpoints; it's asymptotic. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Superb attention to detail on ranges—you're ready for more challenges!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! Composing sin with arcsin undoes the operation, so sin(arcsin(-0.3)) = -0.3, as -0.3 is in the domain [-1,1] and arcsin returns an angle in [-90°,90°] whose sine is -0.3. Choice B correctly evaluates the composition, recognizing that inverse functions cancel each other out. Choice D fails because inverse trig functions do accept negative inputs within their domains, allowing for angles in quadrant IV where sine is negative. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! Arctan(1) = 45°, since tan(45°) = 1 and 45° is within the principal range (-90°, 90°). Choice B correctly identifies the principal value for arctan(1), using the standard reference angle. Choice D fails because 135° is outside the arctan range, even though tan(135°) = -1, not 1, and arctan sticks to its restricted range. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Remembering domain/range restrictions: ARCSIN/ARCTAN: range includes negative angles (quadrant IV) and positive (quadrant I), centered at 0. Makes sense: sine and tangent are negative in quadrant IV, positive in quadrant I. ARCCOS: range is [0°, 180°] (quadrants I and II only, all non-negative angles). Makes sense: cosine is positive in quadrant I, negative in quadrant II, covering all cosine values -1 to 1. Why restrictions needed: sine is periodic (repeats every 360°), so infinitely many angles have sin(θ) = 0.5 (30°, 150°, 390°, -210°, etc.). For arcsin to be a FUNCTION (one output per input), must choose ONE angle to return—convention is the angle in [-90°, 90°] (the "principal value"). This makes arcsin well-defined and usable on calculators!

This question tests your understanding of inverse trigonometric functions (arcsin, arccos, arctan) that "undo" sine, cosine, and tangent to find angles when ratios are known, with restricted domains and ranges making them proper functions. INVERSE TRIGONOMETRIC FUNCTIONS work backwards from regular trig functions: while sin(30°) = 0.5 takes an angle and gives a ratio, arcsin(0.5) = 30° takes a ratio and gives back the angle—it reverses or "undoes" the sine function. The three main inverse functions are: (1) ARCSIN or sin⁻¹: finds angle whose sine is given value, domain is [-1, 1] (possible sine outputs only), range is [-90°, 90°] (quadrants I and IV only—restricted so inverse is a function, since sine isn't one-to-one without restriction). (2) ARCCOS or cos⁻¹: finds angle whose cosine is given value, domain [-1, 1], range [0°, 180°] (quadrants I and II). (3) ARCTAN or tan⁻¹: finds angle whose tangent is given value, domain all real numbers (tangent can be any value), range (-90°, 90°) (not including endpoints, quadrants I and IV). These range restrictions are ESSENTIAL because without them, infinitely many angles have the same sine/cosine value (sin(30°) = sin(150°) = sin(390°) = 0.5), so arcsin must return just ONE answer—it returns the angle in its restricted range! In this triangle, tan(θ) = opposite/adjacent = 5/12, so θ = arctan(5/12) ≈ 22.6°, correctly using arctan since hypotenuse isn't needed. Choice A correctly applies arctan with the right ratio and approximates accurately within (-90°, 90°). A distractor like choice D reverses the ratio to $tan^{-1}$(12/5) ≈67.4°, which finds the complementary angle instead; double-check opposite and adjacent. Using inverse trig to find angles: (1) IDENTIFY which ratio you know: opposite/hypotenuse (use arcsin), adjacent/hypotenuse (use arccos), opposite/adjacent (use arctan). (2) SET UP equation: sin(θ) = ratio → θ = arcsin(ratio). (3) CALCULATE: use calculator arcsin/arccos/arctan button (ensure degree mode if wanting degrees!). (4) CHECK: is answer in expected range? (arcsin gives -90° to 90°, arccos gives 0° to 180°, arctan gives -90° to 90°). If answer outside range, error occurred! Example: right triangle, opposite = 7, hypotenuse = 10. Find angle: sin(θ) = 7/10 = 0.7 → θ = arcsin(0.7) ≈ 44.43°. Check: 44.43° is in range [-90°, 90°] ✓. Terrific job matching functions to ratios—keep practicing!