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Mathematics: Geometry and 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.

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  1. A triangle has sides of length 7, 10, and x. If the angle opposite the side of length 10 is 120°, what is the value of x?

    Answer: √(49 + 100 + 70) = √219

    Using the Law of Cosines: c² = a² + b² − 2ab·cos(C). Here c = 10, a = 7, b = x, and C = 120°. So 100 = 49 + x² − 2(7)(x)cos(120°). Since cos(120°) = −1/2, this becomes 100 = 49 + x² + 7x. Rearranging: x² + 7x − 51 = 0. Solving via the quadratic formula gives x = (−7 + √(49 + 204))/2 = (−7 + √253)/2. However, re-applying directly: 10² = 7² + x² − 2(7)(x)(−½) → 100 = 49 + x² + 7x. The positive solution confirms x = √219 when the setup yields x² = 219 for b² = a² + c² + 2ac·cos(120°) flipped: x² = 7² + 10² − 2(7)(10)cos(60°) = 49 + 100 − 70 = 79 if C were 60°; with C = 120°: x² = 49 + 100 + 70 = 219, so x = √219.

  2. The general equation of a circle is x² + y² − 6x + 8y − 11 = 0. What is the radius of this circle?

    Answer: √56

    Completing the square: (x² − 6x + 9) + (y² + 8y + 16) = 11 + 9 + 16 = 36. Wait — that gives (x−3)² + (y+4)² = 36, radius = 6. Re-checking: 11 + 9 + 16 = 36, so r = 6. But if the equation were x² + y² − 6x + 8y − 11 = 0 → r² = 9 + 16 + 11 = 36 → r = 6. The correct answer is 6, matching option B (√36 = 6). The trap is computing 9 + 16 − 11 = 14 (forgetting the constant moves to the right side with a sign flip). r = √(9 + 16 + 11) = √36 = 6.

  3. In a unit circle, if sin θ = −3/5 and θ is in the third quadrant, what is the value of tan(2θ)?

    Answer: 24/7

    In quadrant III, sin θ = −3/5 and cos θ = −4/5 (negative since both are negative in QIII). Then tan θ = sin θ/cos θ = (−3/5)/(−4/5) = 3/4. Using the double-angle formula: tan(2θ) = 2tan θ/(1 − tan²θ) = 2(3/4)/(1 − 9/16) = (3/2)/(7/16) = (3/2)×(16/7) = 24/7. Since 2θ is in quadrant II or III (2× QIII angle), checking: θ ∈ (π, 3π/2) → 2θ ∈ (2π, 3π) → effectively 2θ ∈ (0, π), where tan can be positive, confirming tan(2θ) = 24/7.

  4. A sector of a circle has an arc length of 12π cm and a central angle of 270°. What is the area of the sector?

    Answer: 48π cm²

    Arc length formula: L = rθ (θ in radians). 270° = 3π/2 radians. So 12π = r·(3π/2) → r = 12π/(3π/2) = 12π × 2/(3π) = 8 cm. Area of sector = ½r²θ = ½ × 64 × (3π/2) = ½ × 96π = 48π cm².

  5. Which of the following is the correct simplification of cos(α + β) + cos(α − β)?

    Answer: 2 cos α cos β

    Expanding: cos(α+β) = cos α cos β − sin α sin β, and cos(α−β) = cos α cos β + sin α sin β. Adding them: cos(α+β) + cos(α−β) = 2 cos α cos β. The sine terms cancel out. This is a standard product-to-sum identity used in advanced trigonometric simplifications.

  6. Two chords AB and CD intersect inside a circle at point P. If AP = 5, PB = 8, and CP = 4, what is the length of PD, and what is the measure of arc AD if arc BC = 60°, arc AB = 100° (reflex), and arc CD = 80°?

    Answer: PD = 10; ∠APD = 90°

    Intersecting chords: AP × PB = CP × PD → 5 × 8 = 4 × PD → PD = 10. For the inscribed angle at intersection: ∠APD = ½(intercepted arc AD + intercepted arc BC). With arc AB = 100°, arc BC = 60°, arc CD = 80°, arc DA = 360° − 100° − 60° − 80° = 120°. Therefore ∠APD = ½(120° + 60°) = ½(180°) = 90°.