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 Find the angle φ.
Show worked solution
sin θ = 6√3 ÷ 12 = √3/2, so θ = 60°. Then φ = 90° − 60° = 30°. Check: cos φ = 6√3/12 = √3/2, and cos 30° = √3/2 ✓. 60° is θ, not φ; 120° = 180° − 60° forgets the right angle; 45° would need two equal short sides.
2 In triangle PQR, the right angle is at Q, PQ = 9 cm and QR = 9√3 cm. Find the angles at P and R.
Show worked solution
tan P = 9√3 ÷ 9 = √3, so angle P = 60°. Angle R = 90° − 60° = 30°. Check: the shortest side, 9, is opposite the smallest angle, R ✓.
3 A right-angled triangle has a hypotenuse of 75 cm, and the side opposite θ is 21 cm. Use the table to find θ, and then give the OTHER acute angle, in degrees.
Show worked solution
sin θ = 21 ÷ 75 = 0.28, so θ = 16° from the table. The other acute angle is 90° − 16° = 74°.
Level 2 · Core
4 A ladder 6√2 m long has its foot 6 m from a vertical wall. What angle does the ladder make with the wall?
Show worked solution
cos(angle with ground) = 6 ÷ 6√2 = 1/√2, so the ladder makes 45° with the ground, and 90° − 45° = 45° with the wall. The height up the wall is also 6 m, so the triangle is isosceles ✓. 135° = 180° − 45° forgets the right angle; 30° and 60° belong to the 1 : √3 : 2 triangle, not this one.
5 A right-angled triangle has sides 16 cm, 30 cm and 34 cm. Given sin 28° ≈ 0.47 and sin 62° ≈ 0.88, find the angle opposite each shorter side.
Show worked solution
sin = 16 ÷ 34 = 8/17 ≈ 0.47, so the angle opposite 16 cm is 28°. sin = 30 ÷ 34 = 15/17 ≈ 0.88, so the angle opposite 30 cm is 62°. Check: 28° + 62° = 90° ✓.
6 The diagonals of a rhombus are 6 cm and 6√3 cm long. Find the obtuse angle of the rhombus, in degrees.
Show worked solution
Each quarter is a right-angled triangle with short sides 3 and 3√3. At the end of the short diagonal, tan = 3√3 ÷ 3 = √3, so the half-angle is 60° and the rhombus angle there is 2 × 60° = 120°. The other angle is 60°, and 120° + 60° = 180° ✓.
Level 3 · Challenge
7 Three statements about an acute angle θ:
1 If cos θ = √3/2, then θ = 30°.
2 If tan θ = √3, the side opposite θ is longer than the hypotenuse.
3 If sin θ = cos θ, then θ = 45°.
Which statements are correct?
1 If cos θ = √3/2, then θ = 30°.
2 If tan θ = √3, the side opposite θ is longer than the hypotenuse.
3 If sin θ = cos θ, then θ = 45°.
Which statements are correct?
Show worked solution
cos 30° = √3/2 (1 ✓). tan θ = √3 gives θ = 60° with sides 1 : √3 : 2, so the opposite side √3 is shorter than the hypotenuse 2; no side is ever longer than the hypotenuse (2 ✗). sin θ = cos θ means O ÷ H = A ÷ H, so O = A and θ = 45° (3 ✓). 1 and 3.
8 The right angle is at C, AC = BC = 6 cm, and D lies on BC with CD = 2√3 cm. Find angle ADB and angle BAD.
Show worked solution
tan(ADC) = 6 ÷ 2√3 = 3/√3 = √3, so angle ADC = 60° and angle ADB = 180° − 60° = 120°. Triangle ABC has AC = BC, so angle B = 45°. In triangle ABD: angle BAD = 180° − 120° − 45° = 15°.
9 An isosceles triangle has a base of 40 cm and a height of 14 cm. Use the table to find the angle at the top, between the two equal sides, in degrees.
Show worked solution
Each half is a right-angled triangle with opposite 14 and adjacent 20, so tan(base angle) = 14 ÷ 20 = 0.7 and the base angle is 35°. The top angle is 180° − 35° − 35° = 110°.