Parallel Lines and Congruence

Angles on a straight line are supplementary, summing to 180°: $180^\circ - 70^\circ = 110^\circ$. (90 − 70 is the complementary angle; 360 − 70 is for angles around a point.)

Angles around a point sum to 360°: $360^\circ - 120^\circ - 150^\circ = 90^\circ$. (Adding gives 270; using 180° is for a straight line.)

Vertically opposite angles are equal, so the angle is $65^\circ$. (115° is the supplementary adjacent angle.)

Complementary angles sum to 90°: $90^\circ - 35^\circ = 55^\circ$. (145° is the supplement: complementary = 90°, supplementary = 180°.)

On parallel lines, corresponding angles (the F-shape) are equal, so $x = 70^\circ$. (110° is the co-interior/supplementary angle.)

Alternate angles (the Z-shape) on parallel lines are equal, so the alternate angle is $65^\circ$. (115° is the supplement.)

Co-interior (C-shape) angles on parallel lines are supplementary: $180^\circ - 80^\circ = 100^\circ$. (Treating them as equal gives 80°.)

Angles in the same position at each intersection (the F-shape) are corresponding angles, equal on parallel lines. (Alternate angles are the Z-shape; co-interior are the C-shape.)

For parallel lines, co-interior angles must sum to exactly 180°. Here $100^\circ + 85^\circ = 185^\circ \ne 180^\circ$, so the lines are not parallel. (Co-interior angles need not be equal; equality is for corresponding and alternate angles.)

Corresponding angles must be equal for lines to be parallel. Here $65^\circ \ne 70^\circ$, so the lines are not parallel.

The interior angles of a triangle sum to 180°. Two obtuse angles would already exceed 180° (more than 90° + more than 90°), leaving nothing for the third angle. So a triangle can have at most one obtuse angle.

$180^\circ - 45^\circ - 65^\circ = 70^\circ$. (Adding the two given angles gives 110°.)

The two equal angles sum to $180^\circ - 80^\circ = 100^\circ$, so each is $\dfrac{100^\circ}{2} = 50^\circ$. (100° forgets to halve.)

By the exterior angle theorem, an exterior angle equals the sum of the two non-adjacent interior angles: $x = 50^\circ + 60^\circ = 110^\circ$. (70° is the interior angle at B.)

The two acute angles of a right triangle are complementary: $90^\circ - 35^\circ = 55^\circ$. (145° forgets to subtract the right angle.)

Substitute $n = 5$: $(5 - 2)\times 180^\circ = 3\times 180^\circ = 540^\circ$. (360° is a quadrilateral; 720° is a hexagon.)

$360^\circ - 90^\circ - 110^\circ - 80^\circ = 80^\circ$. (Adding the three gives 280°.)

Opposite angles in a parallelogram are equal, so the opposite angle is $110^\circ$. (70° is the adjacent, supplementary angle.)

Congruent triangles have the same shape and the same size: all corresponding sides and angles are equal. (Same angles but different size describes similar triangles.)

The valid congruence tests are SSS, SAS, AAS and RHS. AAA is not valid: equal angles only prove similarity (same shape), not congruence (same shape and size).

Angles around a point sum to 360°: $360^\circ - 90^\circ - 130^\circ = 140^\circ$. (Using 180° gives 50°.)

Step 1 (co-interior): $x = 180^\circ - 110^\circ = 70^\circ$. Step 2 (corresponding): $y = x = 70^\circ$.

All three pairs of corresponding sides are equal, so the triangles are congruent by SSS. (SAS needs the included angle; AAA only proves similarity.)

Total parts $2 + 3 + 4 = 9$, so each part is $\dfrac{180^\circ}{9} = 20^\circ$. The largest is 4 parts: $4\times 20^\circ = 80^\circ$. (Check: $40^\circ + 60^\circ + 80^\circ = 180^\circ$.)

Total parts $2 + 3 + 2 + 3 = 10$, so each part is $\dfrac{360^\circ}{10} = 36^\circ$. The smallest is 2 parts: $2\times 36^\circ = 72^\circ$. (Check: $72^\circ + 108^\circ + 72^\circ + 108^\circ = 360^\circ$.)