Problems on Trigonometric Ratios
This topic focuses on problem-solving with trigonometric ratios. Given one trigonometric ratio, you can find all others using the Pythagoras theorem and the definitions of sin, cos, tan, cosec, sec, and cot.
These problems form a significant part of Class 10 board exams. Mastering them requires a clear understanding of how to construct the right triangle, find the missing side, and compute all six ratios.
What is Problems on Trigonometric Ratios?
Trigonometric ratios in a right triangle:
- sin θ = Opposite / Hypotenuse
- cos θ = Adjacent / Hypotenuse
- tan θ = Opposite / Adjacent
- cosec θ = 1/sin θ = Hypotenuse / Opposite
- sec θ = 1/cos θ = Hypotenuse / Adjacent
- cot θ = 1/tan θ = Adjacent / Opposite
sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ | 1 + cot²θ = cosec²θ
Problems on Trigonometric Ratios Formula
Strategy: Given one ratio, find all others.
- From the given ratio, identify the two sides of the right triangle.
- Use Pythagoras theorem to find the third side.
- Write all six ratios using the three sides.
Standard values (memorise):
- sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1
- cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0
- tan 0° = 0, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, tan 90° = undefined
Solved Examples
Example 1: Example 1: Given sin θ, find all ratios
Problem: If sin θ = 3/5, find all other trigonometric ratios.
Solution:
Given: sin θ = Opposite/Hypotenuse = 3/5.
So Opposite = 3k, Hypotenuse = 5k.
- Adjacent = √(5k)² − (3k)² = √(25k² − 9k²) = √(16k²) = 4k
All ratios:
- sin θ = 3/5
- cos θ = 4/5
- tan θ = 3/4
- cosec θ = 5/3
- sec θ = 5/4
- cot θ = 4/3
Answer: cos θ = 4/5, tan θ = 3/4, cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3.
Example 2: Example 2: Given tan θ, find sin and cos
Problem: If tan θ = 12/5, find sin θ and cos θ.
Solution:
Opposite = 12, Adjacent = 5.
- Hypotenuse = √(144 + 25) = √169 = 13
- sin θ = 12/13
- cos θ = 5/13
Answer: sin θ = 12/13, cos θ = 5/13.
Example 3: Example 3: Given cos θ, find tan θ
Problem: If cos θ = 7/25, find tan θ.
Solution:
Adjacent = 7, Hypotenuse = 25.
- Opposite = √(625 − 49) = √576 = 24
- tan θ = 24/7
Answer: tan θ = 24/7.
Example 4: Example 4: Verify an identity
Problem: If sin θ = 8/17, verify that sin²θ + cos²θ = 1.
Solution:
- Opposite = 8, Hypotenuse = 17
- Adjacent = √(289 − 64) = √225 = 15
- cos θ = 15/17
- sin²θ + cos²θ = (8/17)² + (15/17)² = 64/289 + 225/289 = 289/289 = 1 ✓
Answer: Verified. sin²θ + cos²θ = 1.
Example 5: Example 5: Given cosec θ
Problem: If cosec θ = 13/5, find all ratios.
Solution:
cosec θ = Hypotenuse/Opposite = 13/5. So H = 13, O = 5.
- Adjacent = √(169 − 25) = √144 = 12
- sin θ = 5/13, cos θ = 12/13, tan θ = 5/12
- sec θ = 13/12, cot θ = 12/5
Answer: sin = 5/13, cos = 12/13, tan = 5/12, sec = 13/12, cot = 12/5.
Example 6: Example 6: Expression evaluation
Problem: If tan θ = 3/4, evaluate (sin θ + cos θ)/(sin θ − cos θ).
Solution:
- O = 3, A = 4, H = 5
- sin θ = 3/5, cos θ = 4/5
- Numerator = 3/5 + 4/5 = 7/5
- Denominator = 3/5 − 4/5 = −1/5
- Expression = (7/5)/(−1/5) = −7
Answer: −7.
Example 7: Example 7: Using identities directly
Problem: If sin θ = 1/√2, find (1 + tan²θ).
Solution:
sin θ = 1/√2 → θ = 45°. So tan 45° = 1.
- 1 + tan²θ = 1 + 1 = 2
- Alternatively: 1 + tan²θ = sec²θ. cos 45° = 1/√2, sec 45° = √2. sec²45° = 2.
Answer: 2.
Example 8: Example 8: Prove a relationship
Problem: If 3 cos θ = 2, prove that (2sec θ + 3tan θ)/(3sin θ + 2cos θ) has a specific value.
Solution:
cos θ = 2/3. A = 2, H = 3. O = √(9−4) = √5.
- sec θ = 3/2, tan θ = √5/2, sin θ = √5/3
- Numerator = 2(3/2) + 3(√5/2) = 3 + 3√5/2
- Denominator = 3(√5/3) + 2(2/3) = √5 + 4/3
- = (6 + 3√5)/2 ÷ (3√5 + 4)/3 = 3(6 + 3√5) / 2(3√5 + 4)
- = (18 + 9√5) / (6√5 + 8)
This evaluates to a specific numerical value depending on simplification.
Example 9: Example 9: Finding angle from ratio
Problem: If tan θ = √3, find θ (0° ≤ θ ≤ 90°).
Solution:
- tan 60° = √3
- θ = 60°
Answer: θ = 60°.
Example 10: Example 10: Multi-step problem
Problem: If 5 sin θ = 4, find (5 sin θ − 3 cos θ)/(5 sin θ + 2 cos θ).
Solution:
sin θ = 4/5. O = 4, H = 5, A = 3. cos θ = 3/5.
- 5 sin θ = 4, 3 cos θ = 9/5, 2 cos θ = 6/5
- Numerator = 4 − 9/5 = 20/5 − 9/5 = 11/5
- Denominator = 4 + 6/5 = 20/5 + 6/5 = 26/5
- Expression = (11/5)/(26/5) = 11/26
Answer: 11/26.
Real-World Applications
Where these skills are applied:
- Height and distance: Finding heights of buildings, towers using angle of elevation.
- Navigation: Compass bearings and direction calculations.
- Physics: Resolving forces into components, projectile motion.
- Engineering: Slope calculations, ramp design.
- Board exams: 3-4 mark questions on finding ratios are very common.
Key Points to Remember
- Given one ratio, use Pythagoras theorem to find the third side, then compute all six ratios.
- sin²θ + cos²θ = 1 always. Use this to find cos from sin or vice versa.
- tan θ = sin θ / cos θ. cot θ = cos θ / sin θ.
- Memorise the standard values (0°, 30°, 45°, 60°, 90°).
- For expression evaluation: find sin and cos first, then substitute.
- Always take the positive square root for acute angles (0° < θ < 90°).
- Verify by checking sin²θ + cos²θ = 1 after finding the values.
- In exam problems, if given "5 sin θ = 4", rewrite as sin θ = 4/5 first.
Practice Problems
- If cos θ = 5/13, find all six trigonometric ratios.
- If tan θ = 8/15, evaluate (sin θ − cos θ)/(sin θ + cos θ).
- If sec θ = 25/7, find sin θ and tan θ.
- If 3 tan θ = 4, find (4 cos θ − sin θ)/(2 cos θ + sin θ).
- Verify that 1 + tan²θ = sec²θ for tan θ = 7/24.
- If sin θ = √3/2, find θ and all other ratios.
- If cosec θ = 2, find all ratios without finding θ first.
- Evaluate: (1 − sin²30°)(1 + tan²45°).
Frequently Asked Questions
Q1. How do you find all ratios from one given ratio?
Identify two sides from the given ratio, use Pythagoras theorem to find the third side, then write all six ratios using Opposite, Adjacent, and Hypotenuse.
Q2. What if the given ratio involves surds?
Work with surds. For example, if sin θ = √3/2, then cos θ = √(1 − 3/4) = √(1/4) = 1/2.
Q3. Can you use identities instead of Pythagoras?
Yes. sin²θ + cos²θ = 1 directly gives cos θ = √(1 − sin²θ). This avoids constructing a triangle.
Q4. Why do we take the positive root?
For acute angles (0° to 90°), all trigonometric ratios are positive. We only consider acute angles in Class 10.
Q5. What is the fastest way to solve expression problems?
Divide numerator and denominator by cos θ to convert everything to tan θ (if tan is given). This avoids finding sin and cos separately.
Q6. How do I identify which side is opposite, adjacent, or hypotenuse?
Hypotenuse is always opposite the 90° angle (longest side). Opposite is the side across from angle θ. Adjacent is the side next to angle θ (not the hypotenuse).
Q7. What are the reciprocal relationships?
cosec = 1/sin, sec = 1/cos, cot = 1/tan. If sin = 3/5, then cosec = 5/3.
Q8. Can sin θ be greater than 1?
No. For any angle, −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. If a problem gives sin θ > 1, there is no such angle.
