IGCSE Maths · Trigonometry · Core & Extended

Pythagoras’ Theorem and Right-Angled Triangle Trigonometry

This note covers everything Core studies in Trigonometry — Pythagoras and SOHCAHTOA are the whole Core syllabus for this topic. Extended adds perpendicular distance and angles of elevation/depression within the same skillset, tagged below. Everything beyond this note (exact values, graphs, sine/cosine rule, 3D trig) is Extended-only and gets two further notes.

1. Pythagoras’ Theorem

In any right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides.

a² + b² = c²  (c = hypotenuse)
Worked example — finding the hypotenuse 6 cm 8 cm c = ? A right-angled triangle has legs of 6 cm and 8 cm. Find the hypotenuse.
c² = 6² + 8² = 36 + 64 = 100
c = √100 = 10 cm
Worked example — finding a shorter side A right-angled triangle has hypotenuse 13 cm and one leg 5 cm. Find the other leg.
b² = 13² − 5² = 169 − 25 = 144
b = √144 = 12 cm
Try it — Pythagoras’ Theorem
  1. Find the hypotenuse of a right-angled triangle with legs 9 cm and 12 cm.
  2. Find the missing leg of a right-angled triangle with hypotenuse 17 cm and one leg 8 cm.
  3. A rectangle is 24 cm by 7 cm. Find the length of its diagonal.
Show answers
  1. √(81+144) = √225 = 15 cm
  2. √(289−64) = √225 = 15 cm
  3. The diagonal splits the rectangle into two right triangles: √(576+49) = √625 = 25 cm

2. SOHCAHTOA — The Three Trig Ratios

In a right-angled triangle, the three basic ratios connect an angle to the triangle’s sides.

sinθ = opposite / hypotenuse
cosθ = adjacent / hypotenuse
tanθ = opposite / adjacent
Reading the triangle θ opposite adjacent hypotenuse The hypotenuse is always opposite the right angle. The opposite side is across from the angle θ you’re working with; the adjacent side is next to θ (but isn’t the hypotenuse).
Worked example — finding a side A right-angled triangle has an angle of 35° and hypotenuse 20 cm. Find the opposite and adjacent sides.
Opposite = 20 × sin(35°) = 11.5 cm (3 s.f.)
Adjacent = 20 × cos(35°) = 16.4 cm (3 s.f.)

3. Finding Angles Using Inverse Trig Functions

If you know two sides, use the inverse function (sin⁻¹, cos⁻¹, or tan⁻¹ on your calculator) to find the angle.

Worked example A right-angled triangle has an opposite side of 7 cm and an adjacent side of 10 cm. Find the angle θ.
tan θ = 7/10 = 0.7
θ = tan⁻¹(0.7) = 35.0° (1 d.p.)
Try it — SOHCAHTOA
  1. A right-angled triangle has angle 42° and hypotenuse 15 cm. Find the opposite side.
  2. A right-angled triangle has angle 58° and adjacent side 9 cm. Find the hypotenuse.
  3. A right-angled triangle has opposite 12 cm and hypotenuse 20 cm. Find the angle.
  4. A right-angled triangle has adjacent 6 cm and opposite 6 cm. Find the angle (you should be able to guess before calculating).
Show answers
  1. 15×sin(42°) = 10.0 cm
  2. 9/cos(58°) = 17.0 cm
  3. sin⁻¹(12/20) = 36.9°
  4. tan⁻¹(6/6) = tan⁻¹(1) = 45° — since the two legs are equal, this is an isosceles right triangle.

4. Combining Pythagoras and Trigonometry in 2D

Worked example A 5 m ladder leans against a wall, with its base 1.5 m from the wall. Find the angle the ladder makes with the ground, and the height it reaches up the wall.
Angle: cos θ = 1.5/5 = 0.3 → θ = cos⁻¹(0.3) = 72.5° (1 d.p.)
Height (Pythagoras): √(5² − 1.5²) = √22.75 = 4.77 m (3 s.f.)
Try it — Combined Problems
  1. A ramp is 8 m long and rises to a height of 2 m. Find the angle of the ramp with the horizontal.
  2. Using the same ramp, find the horizontal distance it covers.
Show answers
  1. sin⁻¹(2/8) = 14.5°
  2. √(64−4) = √60 = 7.75 m

5. Perpendicular Distance from a Point to a Line Extended only

The shortest distance from a point to a line is always measured along the perpendicular — any other path to the line is longer.

Worked example A point P is 10 m from one end of a straight wall, along a line that makes a 40° angle with the wall. Find the shortest (perpendicular) distance from P to the wall.
The perpendicular distance forms the opposite side of a right-angled triangle, with 10 m as the hypotenuse and 40° as the angle.
Perpendicular distance = 10 × sin(40°) = 6.43 m (3 s.f.)
Try it — Perpendicular Distance Extended
  1. Explain, in one sentence, why the perpendicular distance is always the shortest path from a point to a line.
  2. A point is 14 m from a line, along a path at 55° to the line. Find the perpendicular distance.
Show answers
  1. Any non-perpendicular path forms the hypotenuse of a right triangle with the perpendicular as one leg, and a hypotenuse is always longer than either leg.
  2. 14 × sin(55°) = 11.5 m

6. Angles of Elevation and Depression Extended only

The angle of elevation is measured upward from the horizontal (looking up at something). The angle of depression is measured downward from the horizontal (looking down at something). Because the two horizontal lines involved are parallel, the angle of elevation from one point always equals the angle of depression from the other (alternate angles).

Diagram depression cliff top boat
Worked example From the top of a 50 m cliff, the angle of depression to a boat is 25°. Find the horizontal distance from the base of the cliff to the boat.
The angle of depression (25°) equals the angle of elevation from the boat, so treat this as a right triangle with the 50 m cliff as the opposite side and the horizontal distance as the adjacent side.
tan(25°) = 50 / distance → distance = 50 / tan(25°) = 107 m (3 s.f.)
Try it — Elevation and Depression Extended
  1. From a point on the ground 30 m from a tower’s base, the angle of elevation to the top is 38°. Find the tower’s height.
  2. From an aircraft flying at 2000 m, the angle of depression to an airport is 12°. Find the horizontal distance to the airport.
  3. Two people are 100 m apart. One looks up at a drone at an angle of elevation of 20°, directly between them. Explain how you’d find the drone’s height (you don’t need to fully solve it — just describe the approach).
Show answers
  1. 30 × tan(38°) = 23.4 m
  2. 2000 / tan(12°) = 9410 m (3 s.f.)
  3. Set up a right triangle using the horizontal distance from the person to directly below the drone, and the 20° angle of elevation, then use tan(20°) = height ÷ horizontal distance — though you’d need more information (like where exactly below the drone lies) to pin down the horizontal distance first.
Common mistakes to watch for
  • Mixing up which side is opposite and which is adjacent — this depends on which angle you’re using, not a fixed position in the triangle.
  • Using sin/cos/tan when Pythagoras is needed (or vice versa) — if you only need side lengths and already know two of them, that’s Pythagoras; if an angle is involved, that’s trigonometry.
  • Calculator in the wrong mode (radians instead of degrees) — always check before starting a trig calculation.
  • Confusing angle of elevation and angle of depression — elevation looks up, depression looks down, and they’re measured from the horizontal, not from vertical.
  • Forgetting the perpendicular distance is a specific, shortest-path distance — not just any distance from the point to some part of the line.

7. Quick-Fire Challenge Round

True or False?
  1. Pythagoras’ theorem only works for right-angled triangles.
  2. The hypotenuse is always the shortest side.
  3. The angle of elevation and angle of depression between the same two points are always equal.
  4. Perpendicular distance from a point to a line is always the longest possible distance.
Show answers
  1. True.
  2. False — the hypotenuse is always the longest side.
  3. True. Ext.
  4. False — it’s the shortest possible distance. Ext.
Spot the Error
  1. “Right triangle, legs 5 and 12: hypotenuse = 5+12 = 17.”
  2. “Angle 30°, hypotenuse 10: opposite = 10 × tan(30°).”
  3. “Opposite 8, hypotenuse 17: angle = sin⁻¹(17/8).”
Show answers
  1. Pythagoras requires squaring and adding, not simply adding the sides: √(25+144) = √169 = 13, not 17.
  2. To find the opposite side from the hypotenuse, use sine, not tangent: opposite = 10 × sin(30°) = 5.
  3. The ratio must be opposite ÷ hypotenuse, not the other way round: angle = sin⁻¹(8/17) = 28.1°.

8. Mixed Practice — Bringing It All Together

Core & Extended
  1. Find the hypotenuse of a right triangle with legs 5 cm and 12 cm.
  2. A right triangle has angle 50° and hypotenuse 14 cm. Find the opposite side.
  3. A right triangle has adjacent 9 cm and opposite 11 cm. Find the angle.
Show answers
  1. √(25+144) = 13 cm
  2. 14×sin(50°) = 10.7 cm
  3. tan⁻¹(11/9) = 50.7°
Extended-level mixed set Extended only
  1. A point is 20 m from a line, along a path at 62° to the line. Find the perpendicular distance.
  2. From a lighthouse 45 m tall, the angle of depression to a boat is 18°. Find the distance to the boat.
Show answers
  1. 20×sin(62°) = 17.7 m
  2. 45/tan(18°) = 138.5 m

Practice at Home

A large set again — take it steadily.

Part A — Pythagoras’ theorem
  1. Find the hypotenuse of a right triangle with legs 7 cm and 24 cm.
  2. Find the missing leg of a right triangle with hypotenuse 25 cm and one leg 15 cm.
  3. A ladder reaches 12 m up a wall, with its base 5 m from the wall. Find the ladder’s length.
Part B — SOHCAHTOA, finding sides
  1. A right triangle has angle 28° and hypotenuse 18 cm. Find the opposite and adjacent sides.
  2. A right triangle has angle 63° and opposite side 9 cm. Find the hypotenuse.
  3. A right triangle has angle 47° and adjacent side 11 cm. Find the opposite side.
Part C — SOHCAHTOA, finding angles
  1. A right triangle has opposite 6 cm and hypotenuse 15 cm. Find the angle.
  2. A right triangle has adjacent 10 cm and hypotenuse 22 cm. Find the angle.
  3. A right triangle has opposite 14 cm and adjacent 8 cm. Find the angle.
Part D — Elevation, depression, perpendicular distance Extended
  1. From a point 25 m from a tower’s base, the angle of elevation to the top is 42°. Find the tower’s height.
  2. From the top of a 60 m tall building, the angle of depression to a car is 15°. Find the distance from the building’s base to the car.
  3. A point is 18 m from a line, at an angle of 48° to it. Find the perpendicular distance.
Part E — Applied / challenge
  1. A surveyor stands 40 m from the base of a hill. The angle of elevation to the top is 35°, and the surveyor’s eye level is 1.6 m above the ground. Find the total height of the hill (don’t forget to add the eye height).
  2. Two buildings are 50 m apart. From the top of the shorter building (20 m tall), the angle of elevation to the top of the taller one is 30°. Find the height of the taller building.
  3. A ship sails 15 km due east, then 8 km due north. Find the distance from its starting point (a right-angle bearing problem — Pythagoras applies directly), and the bearing it would need to travel to return directly to the start.

Key Vocabulary Recap

Hypotenuse — the longest side, opposite the right angle
Opposite / adjacent — sides named relative to the angle being used
Angle of elevation — measured upward from the horizontal Ext.
Angle of depression — measured downward from the horizontal Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) sections C6.1 and C6.2 (Core subject content) and sections E6.1 and E6.2 (Extended subject content) — Pythagoras’ theorem and Right-angled triangles. This is the entire Trigonometry syllabus for Core — sections C6.3–C6.6 are explicitly marked “Extended content only.” Perpendicular distance and angles of elevation/depression are Extended-only additions within this shared skillset, marked above.

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