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 KLM, the right angle is at L, KL = 7 and LM = 24. Find cos K.
Show worked solution
KM2 = 72 + 242 = 49 + 576 = 625, so KM = 25. The side next to K is KL = 7, so cos K = \(\dfrac{7}{25}\). \(\dfrac{24}{25}\) is sin K (opposite over hypotenuse); \(\dfrac{7}{24}\) is adjacent over opposite, which is tan M; \(\dfrac{25}{7}\) is upside down.
2 A right-angled triangle has hypotenuse 37 cm and one short side 12 cm. θ is the angle opposite the 12 cm side. Write cos θ as a fraction in simplest form.
Show worked solution
The other short side is √(372 − 122) = √(1369 − 144) = √1225 = 35, and it is next to θ. cos θ = 35/37: numerator 35, denominator 37.
3 In a right-angled triangle, tan θ = 2.4 and the hypotenuse is 39 cm. How long, in cm, is the side opposite θ?
Show worked solution
tan θ = 2.4 = 12/5, so the sides are 12k (opposite), 5k (adjacent) and 13k (hypotenuse, because 52 + 122 = 132). 13k = 39 gives k = 3, so the opposite side is 12 × 3 = 36 cm.
Level 2 · Core
4 A right-angled triangle has acute angles α and β, and cos α = \(\dfrac{20}{29}\). Find tan β.
Show worked solution
The side next to α is 20 and the hypotenuse 29, so the side opposite α is √(841 − 400) = √441 = 21. From β the labels swap: opposite β is 20 and adjacent to β is 21, so tan β = \(\dfrac{20}{21}\). \(\dfrac{21}{20}\) is tan α; \(\dfrac{21}{29}\) is sin α; \(\dfrac{20}{29}\) is cos α, which equals sin β, not tan β.
5 Two right-angled triangles have the same angle θ. In the small one, the side opposite θ is 5 cm and the hypotenuse is 13 cm. The large one has hypotenuse 52 cm. Find the large triangle's other two sides.
Show worked solution
The small adjacent side is √(169 − 25) = 12. The scale factor is 52 ÷ 13 = 4, so the opposite side is 5 × 4 = 20 cm and the adjacent side is 12 × 4 = 48 cm. Check: sin θ = 20/52 = 5/13 ✓.
6 In triangle ABC, the right angle is at B, AB = 21 cm and sin C = 0.28. How long, in cm, is BC?
Show worked solution
AB is opposite C, so 21 ÷ AC = 0.28 and AC = 21 ÷ 0.28 = 75. Then BC2 = 752 − 212 = 5625 − 441 = 5184, so BC = 72 cm. Check: 21, 72, 75 is 3 × (7, 24, 25) ✓.
Level 3 · Challenge
7 In the triangle shown, α and β are the two acute angles.Three statements:
1 sin α = cos β.
2 tan α × tan β = 1.
3 sin α + cos α = 1.
Which statements are correct?
1 sin α = cos β.
2 tan α × tan β = 1.
3 sin α + cos α = 1.
Which statements are correct?
Show worked solution
sin α = 8/17 and cos β = 8/17, because the side opposite α is the side next to β (1 ✓). tan α × tan β = 8/15 × 15/8 = 1 (2 ✓). sin α + cos α = 8/17 + 15/17 = 23/17, which is more than 1 (3 ✗). 1 and 2.
8 In a right-angled triangle, cos θ = \(\dfrac{8}{17}\) and the side opposite θ is 45 cm. Find the other two sides.
Show worked solution
Adjacent 8k and hypotenuse 17k give opposite √(289 − 64)k = 15k. 15k = 45, so k = 3: adjacent 24 cm and hypotenuse 51 cm. Check: 242 + 452 = 576 + 2025 = 2601 = 512 ✓.
9 The right angle is at C, and D lies on BC. AD = 13 cm, AC = 12 cm and tan B = 0.4. How long, in cm, is BD?
Show worked solution
In triangle ACD, CD2 = 132 − 122 = 25, so CD = 5. In triangle ACB, tan B = AC ÷ BC, so 12 ÷ BC = 0.4 and BC = 30. BD = BC − CD = 30 − 5 = 25 cm.