GUJCET Mathematics – Trigonometric Ratios
Prepared for quick revision and deep understanding
1. Simple Explanation
In a right‑angled triangle, the three trigonometric ratios give a relationship between the length of a side and the angle at which that side meets the hypotenuse.
| Ratio | Definition (in terms of a right‑angled triangle) | Symbol |
|---|---|---|
| Sine | Opposite side ÷ Hypotenuse | sin θ |
| Cosine | Adjacent side ÷ Hypotenuse | cos θ |
| Tangent | Opposite side ÷ Adjacent side | tan θ |
The reciprocal ratios are also useful:
| Ratio | Definition | Symbol |
|---|---|---|
| Cosecant | Hypotenuse ÷ Opposite side | csc θ |
| Secant | Hypotenuse ÷ Adjacent side | sec θ |
| Cotangent | Adjacent side ÷ Opposite side | cot θ |
Key Idea – In any right triangle, θ is the angle opposite the side called opposite; the side next to θ (but not the hypotenuse) is adjacent; the longest side is the hypotenuse.
2. Key Points
| # | Concept | Why It Matters for GUJCET |
|---|---|---|
| 1 | Unit‑Circle Representation | All ratios can be read from the unit circle; useful for deriving values at special angles. |
| 2 | Pythagorean Identity | sin²θ + cos²θ = 1. Helps in finding missing ratios. |
| 3 | Reciprocal Identities | csc θ = 1/sin θ, etc. Handy for solving equations where a reciprocal appears. |
| 4 | Angle Addition/Subtraction | sin(A±B), cos(A±B), tan(A±B). Required in problems involving composite angles. |
| 5 | Special Angles | 0°, 30°, 45°, 60°, 90° – memorise their sine, cosine, tangent values. |
| 6 | Rationalizing Denominators | Often the answers are required as simplified fractions or radicals. |
| 7 | Using a Scientific Calculator | GUJCET allows 3‑digit precision; remember to switch between degrees and radians. |
| 8 | Solving Triangles | Law of Sines / Cosines, but for GUJCET primarily only the right‑triangle ratios are needed. |
| 9 | Graphs of Ratios | Knowing the shape of sin x, cos x, tan x (periodic, asymptotes) helps in quick estimation. |
| 10 | Domain & Range | sin, cos ∈ [-1,1]; tan can take any real value except at odd multiples of 90°. |
3. Worked Example
Problem (Typical GUJCET style):
In a right‑angled triangle ABC, ∠B = 30° and the hypotenuse AC = 10 cm.
a) Find the lengths of the other two sides.
b) Computetan ∠Aandcsc ∠B.
Solution
Step 1 – Identify knowns & unknowns
- Angle at B = 30°.
- Hypotenuse
AC = 10. - Opposite side to B = side
BC. - Adjacent side to B = side
AB. - ∠A = 90° – 30° = 60° (since the triangle is right‑angled).
Step 2 – Use sine & cosine
For angle B (30°):
sin 30° = Opposite / Hypotenuse = BC / 10.cos 30° = Adjacent / Hypotenuse = AB / 10.
We know the standard values:
sin 30° = 1/2.cos 30° = √3 / 2.
So,
BC = 10 × 1/2 = 5 cm.AB = 10 × √3 / 2 = 5√3 cm.
Step 3 – Compute tan ∠A
Angle A = 60°.
tan 60° = Opposite / Adjacent relative to angle A.
Here, opposite to A is side BC = 5, adjacent is AB = 5√3.
So,
tan ∠A = BC / AB = 5 / (5√3) = 1/√3 = √3 / 3
Step 4 – Compute csc ∠B
csc θ = 1 / sin θ.
For B = 30°, sin 30° = 1/2.
Hence,
csc ∠B = 1 / (1/2) = 2
Answers
| Quantity | Value |
|---|---|
| BC | 5 cm |
| AB | 5√3 cm |
| tan ∠A | √3 / 3 |
| csc ∠B | 2 |
Quick check – All values satisfy the Pythagorean theorem:
5² + (5√3)² = 25 + 75 = 100 = 10². ✔️
4. Quick Revision Summary
| ✔️ | What to remember |
|---|---|
| 1 | sin θ = opp/hyp, |