Bluebook SAT Test Lines, Angles, and Triangles 6

Question 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?

Accepted Answer

117°

Answer

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°.

Question 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?

Accepted Answer

x = 5

Answer

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° ✓.

Question 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?

Accepted Answer

2

Answer

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.

Question 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?

Accepted Answer

8

Answer

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.

Question 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?

Accepted Answer

6√3

Answer

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 ✓.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Lines, Angles, and Triangles 6