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NBT Mathematics: Trigonometry Flashcards

6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 NBT Mathematics: Trigonometry flashcards as text

  1. Given that sin x = 3/5 with x ∈ (90°, 180°) and cos y = −5/13 with y ∈ (90°, 180°), determine the value of sin(x − y).

    Answer: 33/65

    Since x is in Q2: cos x = −4/5. Since y is in Q2: sin y = +12/13. Using the compound angle identity: sin(x − y) = sin x cos y − cos x sin y = (3/5)(−5/13) − (−4/5)(12/13) = −15/65 + 48/65 = 33/65. The common error is using the wrong sign for cos x or sin y by ignoring the quadrant constraints.

  2. Simplify the expression: [sin(180° + x) · cos(360° − x)] / [tan(180° − x) · sin(90° + x)]

    Answer: cos x

    Apply reduction formulas: sin(180°+x) = −sin x; cos(360°−x) = cos x; tan(180°−x) = −tan x; sin(90°+x) = cos x. Substituting: [(−sin x)(cos x)] / [(−tan x)(cos x)] = (−sin x · cos x) / (−(sin x/cos x) · cos x) = (−sin x · cos x) / (−sin x) = cos x.

  3. How many solutions does the equation sin(2θ) = −√3/2 have for θ ∈ (90°, 270°)?

    Answer: 2

    Let φ = 2θ, so φ ∈ (180°, 540°). Solving sin φ = −√3/2: the reference angle is 60°, giving solutions φ = 240° and φ = 300° within [0°, 360°). In (180°, 540°): φ = 240° and φ = 300° both qualify (the next cycle, φ = 600° and 660°, exceed 540°). This gives θ = 120° and θ = 150°, both in (90°, 270°). So there are exactly 2 solutions.

  4. In triangle PQR, PQ = 7 cm, QR = 9 cm, and PR = 5 cm. Find cos(∠QPR).

    Answer: −1/10

    Using the cosine rule with QR as the side opposite ∠QPR: QR² = PQ² + PR² − 2·PQ·PR·cos(∠QPR). So 81 = 49 + 25 − 2(7)(5)cos(∠QPR) → 81 = 74 − 70cos(∠QPR) → 7 = −70cos(∠QPR) → cos(∠QPR) = −1/10. The negative value indicates ∠QPR is obtuse, which students often miss by misidentifying which side is opposite the target angle.

  5. Which of the following is the simplified form of (sin 3x − sin x) / (cos 3x + cos x)?

    Answer: tan x

    Apply sum-to-product identities: sin 3x − sin x = 2cos((3x+x)/2)sin((3x−x)/2) = 2cos(2x)sin(x). And cos 3x + cos x = 2cos((3x+x)/2)cos((3x−x)/2) = 2cos(2x)cos(x). The expression becomes [2cos(2x)sin(x)] / [2cos(2x)cos(x)] = sin x/cos x = tan x. The distractor tan(2x) tempts students who misapply the formulas.

  6. From point A on level ground, the angle of elevation to the top P of a vertical pole PQ is α, where tan α = 1/2. From point B on the same line, 10 m closer to the base Q, the angle of elevation is β, where tan β = 1. What is the height of the pole PQ?

    Answer: 10 m

    Let h = PQ and d = AQ. From A: tan α = h/d = 1/2, so d = 2h. From B: BQ = d − 10 = 2h − 10. Applying tan β = h/BQ = 1 gives h = 2h − 10, so h = 10 m. Students who set up BQ incorrectly or confuse which distance decreases often get 20 m or 5 m.