Angle facts and parallel lines

The marked angle is bigger than $180^\circ$ but less than $360^\circ$, so it is a reflex angle.

Place the centre of the protractor on the vertex and line zero up with the vertical arm.

Read from the scale that starts at 0 on that arm: about $40^\circ$.

(a) Angles around a point add up to $360^\circ$.

$x = 360 - 50 = 310^\circ$

(b) An obtuse angle is between $90^\circ$ and $180^\circ$. $50^\circ$ is less than $90^\circ$, so it is an acute angle.

(a) The two angles go all the way round the point.

$y = 360 - 70 = 290$

(b) Angles at a point add up to $360°$.

(i) Angles on a straight line total $180^\circ$.

$x = 180 - 75 - 84 = 21$

(ii) Angles on a straight line add up to $180^\circ$.

Add the angles: $23 + 145 + 23 + 69 = 260°$.

Angles around a point must add up to $360°$, so these results cannot be right.

(a) Angles on a straight line add up to $180^\circ$, so $x = 180 - 150 = 30^\circ$.

(b) Angles around a point make $360^\circ$. Here $90 + 280 = 370^\circ$, which is too big (the reflex angle should be $270^\circ$).

Angles on a straight line add up to $180°$.

$x = 180 - 100 - 35 = 45°$

Angles at a point add up to $360^\circ$.

$x = 360 - 220 - 90 = 50^\circ$

Angle $ACB = 180 - 116 - 25 = 39^\circ$ (angles in a triangle add up to $180^\circ$).

$x$ and angle $ACB$ are vertically opposite, so $x = 39^\circ$ (vertically opposite angles are equal).

(a) The angle at B is a right angle, so $x = 90 - 25 - 25 = 40$

(b)(i) $b = 125°$ because vertically opposite angles are equal (or $d = 125°$ because corresponding angles are equal).

(ii) Angles around a point add up to $360°$, so $a + b + c = 360 - 125 = 235°$

Angle $BAC = 360 - 310 = 50^\circ$ (angles around a point add to $360^\circ$).

Angle $ACB = 180 - 115 = 65^\circ$ (angles on a straight line add to $180^\circ$).

Angle $ABC = 180 - 50 - 65 = 65^\circ$ (angles in a triangle add to $180^\circ$).

Angles $ACB$ and $ABC$ are both $65^\circ$, so the triangle is isosceles.

Angle $AEB = 63°$ (corresponding angles are equal).

Angle $ABE = 180 - 148 = 32°$ (angles on a straight line add up to $180°$).

Angle $EAB = 180 - 63 - 32 = 85°$ (angles in a triangle add up to $180°$).

Triangle ABD is isosceles ($AB = BD$), so $\angle ADB = 64°$ (base angles of an isosceles triangle are equal).

$\angle ABD = 180 - 64 - 64 = 52°$ (angles in a triangle add up to $180°$).

$\angle DBC = 180 - 52 = 128°$ (angles on a straight line add up to $180°$).

Triangle DBC is isosceles ($BD = BC$), so $x = \frac{180 - 128}{2} = 26°$

(a) The marked equal sides meet at C, so the base angles are at A and B. Angle A $= 63°$, so $x = 180 - 63 - 63 = 54°$.

(b) The 57° angle and angle EGH lie on opposite sides of the line EG, between the parallel lines. That makes them alternate angles, not corresponding.

The two marked angles add up to $90°$ because the third angle is a right angle.

$7x + 5x + 18 = 90$, so $12x = 72$ and $x = 6$.

The angles are $7 \times 6 = 42°$, $5 \times 6 + 18 = 48°$ and $90°$, so the smallest is $42°$.

Angle $A = 180 - 90 - 22 = 68°$ (angles in a triangle add to $180°$)

Triangle APQ is isosceles since AP = AQ, so angles AQP and APQ are equal (base angles of an isosceles triangle).

Angle $AQP = (180 - 68) \div 2 = 56°$

AB is parallel to DC, so angle $ACD = e$ (alternate angles are equal).

Angles in triangle $ACD$ add up to $180°$:

angle $CAD = 180 - 3e - e = 180 - 4e$

BA = BD, so the angles at A and D are equal: both are $x$.

Angles in the triangle: $x + x + y = 180$. With $x = 2y$: $5y = 180$, so $y = 36$.

Angles on a straight line: $w = 180 - 36 = 144$

Triangle BCD is equilateral, so angle $DBC = 60°$ (angles in an equilateral triangle are equal).

Angle $BDE = 60°$ (alternate angles are equal, since CB is parallel to DE).

Angle $ABD = 90°$ (marked right angle).

Angles in a quadrilateral add up to $360°$: $x = 360 - 90 - 60 - 132 = 78°$

Angle $CFG = 180 - 125 = 55°$ (angles on a straight line add to 180°).

Angle $ACD = 55°$ (corresponding angles are equal).

Angle $ADC = 180 - 110 = 70°$ (angles on a straight line).

Angle $CAD = 180 - 55 - 70 = 55°$ (angles in a triangle add to 180°).

Two angles are equal, so triangle $ACD$ is isosceles.

Angle $QPR = 56°$ (vertically opposite angles are equal).

Angle $PQR = (180 - 56) \div 2 = 62°$ (base angles of an isosceles triangle are equal; angles in a triangle add to 180°).

Angle $PRQ = 62°$, and $CQD \parallel APR$, so angle $CQR = 180 - 62 = 118°$ (co-interior angles add to 180°).

Angle $ACB = 180 - 75 - 51 = 54°$

Split 54 in the ratio 2 : 1: one part $= 54 \div 3 = 18°$

So $ACD = 18°$ and $DCB = 36°$

In triangle $BDC$: angle $BDC = 180 - 51 - 36 = 93°$

Angle $AOB$ is not obtuse, so $2x + 3x + 10 \le 90$

$5x \le 80$

$x \le 16$, so the greatest value is $x = 16$

Angle $BAD = 100°$ (marked on the diagram).

Angle $ADC = 180 - 75 = 105°$ (angles on a straight line add up to 180°).

Angle $BCD = 50°$ (vertically opposite angles are equal).

Angle $ABC = 360 - 100 - 50 - 105 = 105°$ (angles in a quadrilateral add up to 360°).

$AB = AD$, so triangle $ABD$ is isosceles and angles $ABD$ and $ADB$ are equal. Since angle $ABC$ = angle $ADC$, angles $CBD$ and $CDB$ are also equal, so $CB = CD$.

Two pairs of equal adjacent sides, so $ABCD$ is a kite.

Angle $ABE = 180 - 110 = 70°$ (co-interior angles add up to 180°)

Angle $EBG = 180 - 70 - 35 = 75°$ (angles on a straight line add up to 180°)

Angle $BEG = 180 - 110 - 25 = 45°$ (angles on a straight line)

$x = 180 - 75 - 45 = 60°$ (angles in a triangle add up to 180°)

(a) If A is one of the two equal angles and B is the other: $B = 70^\circ$.

If A and C are equal: $B = 180 - 70 - 70 = 40^\circ$.

If A is the odd angle: $B = (180 - 70) \div 2 = 55^\circ$.

(b) Two angles of $120^\circ$ would add to more than $180^\circ$, so 120° must be the single different angle. Then $B = (180 - 120) \div 2 = 30^\circ$ only.

Angles in a quadrilateral add to $360^\circ$:

$(2x+15) + (4x+15) + (4x+8) + (3x-3) = 360$

$13x + 35 = 360 \Rightarrow x = 25$

$A = 65^\circ$, $B = 115^\circ$, so $A + B = 180^\circ$.

Co-interior angles add to $180^\circ$, so $AD$ is parallel to $BC$ and $ABCD$ is a trapezium.

In triangle ABC, $AB = BC$ so angle BCA = angle CAB = $81°$.

Angle ABC $= 180 - 2 \times 81 = 18°$

Angle BCD $= 4 \times 18 = 72°$

In triangle BCD, $BC = CD$ so angle CBD = angle CDB $= (180 - 72) \div 2 = 54°$

$18 : 54 = 1 : 3$