Lines, Angles, and Triangles Flashcards

In triangle ABC, the exterior angle at vertex C measures 112°. If angle A is 15° more than angle B, what is the measure of angle B?

Answer: 38.5°

The exterior angle of a triangle equals the sum of the two non-adjacent interior angles, so angle A + angle B = 112°. Since angle A = angle B + 15°, substituting gives (angle B + 15°) + angle B = 112°, so 2·angle B = 97°, meaning angle B = 48.5°. Wait — re-checking: angle A + angle B = 112°, angle A = angle B + 15°, so angle B + 15 + angle B = 112, 2·angle B = 97, angle B = 48.5°. But the answer choices show 38.5°. Let me reread: angle A is 15° MORE than angle B means A = B + 15. A + B = 112 → (B+15)+B = 112 → 2B = 97 → B = 48.5°. The correct answer is 48.5°.

Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle is expressed as (5x − 18)° and the other as (3x + 6)°. A third line crosses both parallel lines creating an angle of (2x + 4)° with the transversal at one intersection. What is the value of x?

Answer: 21

Co-interior (consecutive interior) angles formed by a transversal cutting two parallel lines are supplementary, summing to 180°. So (5x − 18) + (3x + 6) = 180 → 8x − 12 = 180 → 8x = 192 → x = 24. The third line information is a distractor — x = 24 is fully determined by the co-interior angle relationship alone.

In triangle PQR, PQ = PR (isosceles) and angle QPR = 4k°. A line segment from P bisects angle QPR and meets QR at point S. If QS = 7 and QR = 18, what is the value of angle PQR in terms of k?

Answer: (90 − 2k) degrees

Since PQ = PR, triangle PQR is isosceles, so the base angles are equal: angle PQR = angle PRQ. The three angles must sum to 180°, so 2·(angle PQR) + 4k° = 180°, giving angle PQR = (180° − 4k°)/2 = 90° − 2k°. The segment lengths QS and QR are distractors — they don't change the angle relationship. The answer is (90 − 2k)°.

Line ℓ₁ passes through (−3, 7) and (5, −1). Line ℓ₂ is perpendicular to ℓ₁ and passes through (4, 2). At what point do ℓ₁ and ℓ₂ intersect?

Answer: (3, −0.5)

Slope of ℓ₁ = (−1 − 7)/(5 − (−3)) = −8/8 = −1. Perpendicular slope = 1. Line ℓ₂: y − 2 = 1·(x − 4) → y = x − 2. Line ℓ₁: y − 7 = −1·(x + 3) → y = −x + 4. Setting equal: x − 2 = −x + 4 → 2x = 6 → x = 3, y = 1. The intersection is (3, 1).

In triangle XYZ, a cevian XW divides YZ such that YW:WZ = 2:3. The area of triangle XYW is 24 cm². What is the area of triangle XYZ?

Answer: 60 cm²

A cevian from X to side YZ creates two triangles (XYW and XWZ) that share the same height from X. Their areas are proportional to their bases YW and WZ. Since YW:WZ = 2:3, the area of XWZ = (3/2) × 24 = 36 cm². The total area of XYZ = 24 + 36 = 60 cm².

Two lines intersect forming four angles. One angle is (7m − 11)° and the angle that is neither adjacent nor vertical to it is (4m + 31)°. What is the measure of the largest angle formed at the intersection?

Answer: 100°

The angle that is neither adjacent nor vertical to a given angle at an intersection IS the vertical angle — there are only two distinct pairs at any intersection. Vertical angles are equal, so 7m − 11 = 4m + 31 → 3m = 42 → m = 14. The angle = 7(14) − 11 = 98 − 11 = 87°. Its supplement (adjacent angle) = 180° − 87° = 93°. Wait — re-examining: the problem says 'neither adjacent nor vertical,' which at a two-line intersection is impossible (every angle is either adjacent or vertical to any other). This is a trick — 'neither adjacent nor vertical' describes no angle, so the two given expressions must be supplementary (adjacent). 7m − 11 + 4m + 31 = 180 → 11m + 20 = 180 → 11m = 160 → m = 160/11. That doesn't work cleanly. Setting them as vertical: m=14, angles = 87° and 93°. Largest = 93°. Closest answer is 100°. The intended reading is that the two given angles are supplementary: 7m−11+4m+31=180 → 11m=160, not clean. So they're vertical: m=14, giving 87° and supplement 93°. The largest angle is 93°.