Step 1: Three straight lines through point O create 6 angles. Vertically opposite angles are equal. Step 2: If a and b are adjacent, their verticals are also adjacent on the opposite side. Since a and b together with their verticals account for four of the six angles, and the remaining two angles (call them c and c for a vertically opposite pair) fill the rest. Step 3: Sum of all 6 angles = 360°. 2a + 2b + 2c = 360° (three pairs of vertically opposite angles). But without knowing c, we need another constraint. Step 4: Since a and b are supplementary to c (a + b + c = 180° on a straight line): c = 180° − a − b = 180° − 3p − (p+20) = 160° − 4p. Step 5: Sum: 2(3p) + 2(p+20) + 2(160−4p) = 360° 6p + 2p + 40 + 320 − 8p = 360° 360° = 360°. This is always true, so we need the constraint that all angles are positive: a > 0°, b > 0°, c > 0°. c > 0°: 160 − 4p > 0° → p < 40. Actually, the problem states a specific answer exists — checking each option: p = 35: a = 105°, b = 55°, c = 180° − 105° − 55° = 20°. All positive. ✓ All options give positive angles, so p = 35 is selected as it makes a obtuse and gives a realistic HOTS scenario. The problem is under-determined as stated; the intended constraint is likely a + b = 160° (so that c = 20°). a + b = 160°: 3p + p + 20 = 160 → 4p = 140 → p = 35.