Angles & angle relationships

Fluency

acute
right
obtuse
reflex
straight
$x = 102^\circ$
$x = 61^\circ$
$x = 230^\circ$ (the two angles at a point sum to $360^\circ$ )
$x = 120^\circ$
$63^\circ$ (vertically opposite); two of $117^\circ$ (on the line with the $63^\circ$ )
$80^\circ$
$50^\circ$ each
$55^\circ$
$x = 73^\circ$
$x = 112^\circ$
$x = 72^\circ$

Reasoning

$x = 40^\circ$ . Method: $x + 2x + 60 = 180$ , so $3x = 120$ .
$x = 85^\circ$ . Method: $x + x + 40 + 150 = 360$ , so $2x = 170$ .
$30^\circ, 60^\circ, 90^\circ$ . Method: $1 + 2 + 3 = 6$ parts; each part $= 30^\circ$ .
$x = 20^\circ$ ; angles $40^\circ, 60^\circ, 80^\circ$ . Method: $9x = 180$ .
$120^\circ$ . Reason: exterior angle equals the sum of the two non-adjacent interior angles.
$36^\circ$ . Method: both base angles are $72^\circ$ ; apex $= 180 - 144$ .
$x = 22.5^\circ$ . Method: $3x + 5x = 180$ .
$x = 25^\circ$ . Method: $x + 2x + 15 = 90$ , so $3x = 75$ .
$x = 83\tfrac{1}{3}^\circ$ . Method: $3x + 110 = 360$ , so $3x = 250$ .
$x = 50^\circ$ . Method: corresponding angles are equal, so $2x + 20 = x + 70$ , hence $x = 50$ .

Reasoning

Not always true. Vertically opposite angles are equal, not supplementary. They only add to $180^\circ$ in the special case where both are $90^\circ$ . The pair that sums to $180^\circ$ is the pair of angles on a straight line (adjacent angles at the crossing), not the vertically opposite pair.
Emma is essentially correct: any three positive angles that sum to $180^\circ$ can be the angles of some triangle. The caveat is that each angle must be positive - e.g. $0^\circ, 0^\circ, 180^\circ$ sums to $180^\circ$ but cannot form a triangle.
Tom is wrong. Co-interior angles are supplementary (sum to $180^\circ$ ), not equal. He is confusing co-interior with alternate or corresponding angles, which are equal on parallel lines.
Not possible. The three angles in a triangle must sum to $180^\circ$ . Two right angles already account for $180^\circ$ , leaving $0^\circ$ for the third - which is not a valid angle in a triangle.

Problem solving

$90^\circ$ . The $12$ and $3$ positions form a right angle.
$180^\circ$ . The hands point in opposite directions.
$55^\circ$ with the wall. Method: wall and floor are perpendicular; $90 - 35$ .
$75^\circ$ with the ground. Method: $90 - 15$ .
The four angles are $48^\circ$ , $132^\circ$ , $48^\circ$ , $132^\circ$ . The acute $48^\circ$ and its vertically opposite pair give one set; the other two are $180 - 48 = 132^\circ$ each.
$37^\circ$ . Method: $180 - 90 - 53$ .

Tier 1: basic skills