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Lines, Angles, and Triangles Flashcards

6 cards from real Bluebook SAT Test 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. In triangle ABC, m∠A = 54°. The angle bisectors of ∠B and ∠C intersect at the incenter I. What is the measure of ∠BIC?

    Answer: 117°

    The incenter angle formula states that ∠BIC = 90° + (∠A / 2). Substituting: ∠BIC = 90° + (54° / 2) = 90° + 27° = 117°. This follows from the fact that in △BIC, the angles at B and C are each half of ∠B and ∠C respectively, and their sum plus ∠BIC equals 180°.

  2. In a triangle, two interior angles measure (x² + 2x)° and (x + 10)°. The exterior angle at the third vertex measures (4x + 30)°. If x > 0, what is the value of x?

    Answer: x = 5

    The Exterior Angle Theorem states that an exterior angle equals the sum of the two non-adjacent interior angles: 4x + 30 = (x² + 2x) + (x + 10). Simplifying: 4x + 30 = x² + 3x + 10, which gives x² − x − 20 = 0, or (x − 5)(x + 4) = 0. The solutions are x = 5 and x = −4. Since x > 0, x = 5. Verification: angles are 35°, 15°, and 130°; exterior angle = 50° = 35° + 15° ✓.

  3. A triangle has sides of length (3x − 2), (x + 7), and (2x + 3). What is the minimum integer value of x for which a valid triangle can be formed?

    Answer: 2

    All three triangle inequalities must hold. (1) (3x−2)+(x+7) > (2x+3) → x > −1. (2) (3x−2)+(2x+3) > (x+7) → 4x > 6 → x > 1.5. (3) (x+7)+(2x+3) > (3x−2) → 10 > −2, always true. Also, all sides must be positive: 3x−2 > 0 requires x > 2/3. The strictest constraint is x > 1.5, so the minimum integer value is x = 2.

  4. In triangle ABC, segment BD bisects ∠B and meets AC at point D. If AB = 10, BC = 15, and AC = 20, what is the length of AD?

    Answer: 8

    By the Angle Bisector Theorem, the bisector divides the opposite side in the ratio of the two adjacent sides: AD/DC = AB/BC = 10/15 = 2/3. Since AD + DC = 20, we get AD = (2/5) × 20 = 8. A common error is using BC/AB = 3/2, which gives AD = 12 — but the ratio is AB:BC, not BC:AB.

  5. In right triangle ABC with the right angle at C, altitude CD is drawn to hypotenuse AB. If AD = 3 and AB = 12, what is the length of BC?

    Answer: 6√3

    When the altitude is drawn to the hypotenuse, each leg is the geometric mean of the full hypotenuse and the adjacent hypotenuse segment. So BC² = AB × DB. Since DB = AB − AD = 12 − 3 = 9, we get BC² = 12 × 9 = 108, giving BC = √108 = 6√3. Verification: AC² = AB × AD = 12 × 3 = 36, so AC = 6, and AC² + BC² = 36 + 108 = 144 = AB² ✓.

  6. In triangle ABC, m∠A = 60°. The exterior angle at vertex B is exactly twice the exterior angle at vertex C. What is m∠B?

    Answer: 20°

    Let the exterior angle at C = e°, so the exterior angle at B = 2e°. The corresponding interior angles are (180 − 2e)° at B and (180 − e)° at C. Applying the triangle angle sum: 60 + (180 − 2e) + (180 − e) = 180 → 420 − 3e = 180 → e = 80°. Therefore m∠B = 180° − 2(80°) = 20°. Check: m∠C = 180° − 80° = 100°, and 20° + 100° + 60° = 180° ✓.