Trigonometric ratios of an acute angle
In a right triangle, each trigonometric ratio compares two sides as seen from an acute angle. The ratio depends only on the angle, not the triangle's size.
Naming the sides
Trigonometry means measuring triangles. Take a triangle ABC with ∠B = 90°. Look at the sides from angle A. BC is the opposite side: it faces angle A. AB is the adjacent side: it touches angle A. AC is the hypotenuse: the longest side, facing the right angle. Many teachers call the opposite side the perpendicular (P) and the adjacent side the base (B). If you look from angle C instead, opposite and adjacent swap, but the hypotenuse stays. Right triangles with the same angle A are similar, so their sides are in the same ratio. So sin A depends only on angle A. Since the hypotenuse is the longest side, sin A and cos A are less than 1 for an acute angle. Note that sin A is one symbol, not sin × A.
The six trigonometric ratios of angle A (ΔABC, ∠B = 90°)
| Ratio | Short form | Sides |
|---|---|---|
| sine of A | sin A | opposite/hypotenuse = BC/AC |
| cosine of A | cos A | adjacent/hypotenuse = AB/AC |
| tangent of A | tan A | opposite/adjacent = BC/AB |
| cosecant of A | cosec A | hypotenuse/opposite = AC/BC |
| secant of A | sec A | hypotenuse/adjacent = AC/AB |
| cotangent of A | cot A | adjacent/opposite = AB/BC |
How to draw the figure for a ratio question
- Draw a right triangle ABC and mark the right angle at B with a small square.
- Mark the angle you are asked about (A or θ) with a small arc.
- Write hypotenuse along AC, the side facing the right angle.
- For angle A, label BC as opposite and AB as adjacent.
- Write the given lengths on their sides with units, and call the unknown side x.
Finding all six ratios
In ΔABC, right-angled at B, AB = 8 cm and BC = 15 cm. Find the six trigonometric ratios of angle A. Also find sin C and cos C.
- By Pythagoras theorem, AC² = AB² + BC².
- AC² = 8² + 15² = 64 + 225 = 289, so AC = 17 cm.
- For angle A: opposite = BC = 15, adjacent = AB = 8, hypotenuse = AC = 17.
- sin A = 15/17, cos A = 8/17, tan A = 15/8.
- cosec A = 17/15, sec A = 17/8, cot A = 8/15.
- For angle C: opposite = AB = 8, adjacent = BC = 15.
- sin C = 8/17, cos C = 15/17.
sin A = 15/17, cos A = 8/17, tan A = 15/8, cosec A = 17/15, sec A = 17/8, cot A = 8/15; sin C = 8/17, cos C = 15/17.
Word problem: a ladder against a wall
A 13 m long ladder leans against a wall. Its foot is 5 m from the wall. θ is the angle between the ladder and the ground. Find sin θ, cos θ and tan θ.
- The wall, the ground and the ladder form a right triangle. The ladder (13 m) is the hypotenuse.
- The side adjacent to θ is the ground distance, 5 m.
- Height on the wall (opposite side) = √(13² − 5²) = √(169 − 25) = √144 = 12 m.
- sin θ = opposite/hypotenuse = 12/13.
- cos θ = adjacent/hypotenuse = 5/13.
- tan θ = opposite/adjacent = 12/5.
sin θ = 12/13, cos θ = 5/13, tan θ = 12/5.
A carpenter fixing a wheelchair ramp, or a kite flyer on Makar Sankranti judging the slope of the string, is using a ratio of sides. The steeper the angle, the larger tan θ.
1 marks: “In ΔABC, right-angled at B, AB = 20 cm and BC = 21 cm. Find sin A.” AC = √(400 + 441) = √841 = 29 cm. sin A = BC/AC = 21/29.
2 marks: “In ΔPQR, right-angled at Q, PQ = 9 cm and PR = 41 cm. Find sin R, cos R and tan R.” QR = √(1681 − 81) = √1600 = 40 cm. For angle R, opposite = PQ = 9. sin R = 9/41, cos R = 40/41, tan R = 9/40.
3 marks: “In ΔXYZ, right-angled at Y, XY = 12 cm and YZ = 5 cm. Find all six trigonometric ratios of angle Z.” XZ = √(144 + 25) = 13 cm. For angle Z, opposite = XY = 12, adjacent = YZ = 5. sin Z = 12/13, cos Z = 5/13, tan Z = 12/5, cosec Z = 13/12, sec Z = 13/5, cot Z = 5/12.
Pandit Badri Prasad Har Har Bole: sin = P/H, cos = B/H, tan = P/B. P is the perpendicular (opposite), B the base (adjacent), H the hypotenuse.