Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 In triangle PQR, PQ = PR. Side QR is extended to S, and ∠PRS = 128°. Find ∠QPR.
Show worked solution
∠PRQ = 180 − 128 = 52°, and PQ = PR makes ∠PQR = 52° too. So ∠QPR = 180 − 104 = 76°. (Or by the exterior angle theorem: 128 = ∠QPR + 52.) 52° is only a base angle; 104° adds the base angles but never takes them from 180°; 64° halves 128°, as if ∠QPR and ∠PQR were equal.
2 Each interior angle of a regular polygon is 162°. Find the number of sides and the sum of the interior angles.
Show worked solution
Exterior angle = 180 − 162 = 18°, so n = 360 ÷ 18 = 20 sides. Angle sum = (20 − 2) × 180 = 3240°.
3 The two horizontal lines are parallel. Find x, in degrees.
Show worked solution
Vertically opposite the 124° is the angle below the top line on the left of the transversal, also 124°. That angle and x lie between the parallel lines on the same side of the transversal, so they are co-interior: x = 180 − 124 = 56°.
Level 2 · Core
4 A five-pointed star has tip angles 29°, 37° and 46°, and its other two tips are equal, each y. Find y.
Show worked solution
The tips add to 180°: 2y = 180 − (29 + 37 + 46) = 180 − 112 = 68, so y = 34°. 68° forgets to halve; 124° uses 360° as the total, (360 − 112) ÷ 2; 214° uses the 540° of the pentagon in the middle, (540 − 112) ÷ 2.
5 B, D and C lie on a straight line, BD = AD = AC and ∠ABD = 35°. Find ∠ADC and ∠BAC.
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BD = AD, so ∠BAD = 35° and ∠ADC = 35 + 35 = 70° (exterior angle of triangle ABD). AD = AC, so ∠ACD = 70° and ∠DAC = 180 − 140 = 40°. So ∠BAC = 35 + 40 = 75°.
6 The interior angles of a pentagon are x°, (x + 10)°, (x + 20)°, (x + 30)° and (x + 40)°. Find the largest angle, in degrees.
Show worked solution
A pentagon's angles add to (5 − 2) × 180 = 540°. So 5x + 100 = 540, 5x = 440 and x = 88. The largest angle is 88 + 40 = 128°.
Level 3 · Challenge
7 Three statements:
1 A triangle with angles of 50° and 65° is isosceles.
2 Three regular pentagons can meet at a point with no gap.
3 A regular polygon with interior angle 140° has 9 sides.
Which statements are correct?
1 A triangle with angles of 50° and 65° is isosceles.
2 Three regular pentagons can meet at a point with no gap.
3 A regular polygon with interior angle 140° has 9 sides.
Which statements are correct?
Show worked solution
The third angle is 180 − 115 = 65°, so two angles are equal and the triangle is isosceles (1 ✓). 3 × 108 = 324°, which leaves a 36° gap (2 ✗). The exterior angle is 40°, and 360 ÷ 40 = 9 (3 ✓). 1 and 3.
8 ABCDE is a regular pentagon, and ABP is an equilateral triangle drawn inside it. Find ∠BPC and ∠CPE.
Show worked solution
∠PBC = 108 − 60 = 48°, and BP = BA = BC, so triangle PBC is isosceles: ∠BPC = (180 − 48) ÷ 2 = 66°. By symmetry ∠APE = 66° too. The angles around P add to 360°, so ∠CPE = 360 − 60 − 66 − 66 = 168°.
9 The two horizontal lines are parallel. Find x, in degrees.
Show worked solution
Through C draw a parallel line: the 75° splits into 28° (alternate to the angle at D) and 47°. Through B draw another: the lower part of x is alternate to that 47°, and the upper part is alternate to the 32° at A. So x = 32 + 47 = 79°. Check: x + 28 = 32 + 75 = 107.