Trigonometry in Right Triangles
Trigonometry originated from the study of right triangles. The trigonometric ratios — sin, cos, tan — are defined as ratios of the sides of a right-angled triangle with respect to one of its acute angles.
This topic focuses on applying trigonometric ratios to solve right triangles — finding unknown sides and angles when some information is given.
These skills are directly used in height and distance problems, navigation, engineering, and physics.
What is Trigonometry in Right Triangles?
In a right triangle with acute angle θ:
- sin θ = Side opposite to θ / Hypotenuse
- cos θ = Side adjacent to θ / Hypotenuse
- tan θ = Side opposite to θ / Side adjacent to θ
Solving a right triangle means finding all unknown sides and angles given some initial information.
SOH-CAH-TOA
Sin = Opposite/Hypotenuse | Cos = Adjacent/Hypotenuse | Tan = Opposite/Adjacent
Trigonometry in Right Triangles Formula
To find a side:
- If you know an angle and the hypotenuse: use sin or cos.
- If you know an angle and one leg: use tan, sin, or cos.
To find an angle:
- If you know two sides: use the inverse ratio (e.g., θ = tan⁻¹(opposite/adjacent)).
- At Class 10 level, identify the angle from standard values (30°, 45°, 60°).
Pythagoras theorem is always available: a² + b² = c² (where c is the hypotenuse).
Solved Examples
Example 1: Example 1: Finding a side using sin
Problem: In a right triangle, the hypotenuse is 20 cm and one angle is 30°. Find the side opposite to 30°.
Solution:
- sin 30° = Opposite / Hypotenuse
- 1/2 = Opposite / 20
- Opposite = 10 cm
Answer: The opposite side = 10 cm.
Example 2: Example 2: Finding a side using cos
Problem: In a right triangle, the hypotenuse is 14 cm and one angle is 60°. Find the adjacent side.
Solution:
- cos 60° = Adjacent / Hypotenuse
- 1/2 = Adjacent / 14
- Adjacent = 7 cm
Answer: The adjacent side = 7 cm.
Example 3: Example 3: Finding a side using tan
Problem: A ladder leans against a wall making 60° with the ground. If the foot is 5 m from the wall, find the height reached by the ladder.
Solution:
- tan 60° = Height / Base
- √3 = Height / 5
- Height = 5√3 ≈ 8.66 m
Answer: Height = 5√3 ≈ 8.66 m.
Example 4: Example 4: Finding the hypotenuse
Problem: A ramp makes 30° with the ground and rises 3 m vertically. Find the length of the ramp.
Solution:
- sin 30° = Opposite / Hypotenuse = 3 / Ramp
- 1/2 = 3 / Ramp
- Ramp = 6 m
Answer: Ramp length = 6 m.
Example 5: Example 5: Finding an angle
Problem: In a right triangle, the two legs are 7 cm and 7 cm. Find the acute angles.
Solution:
- tan θ = 7/7 = 1
- θ = 45°
- Both acute angles are 45° (isosceles right triangle).
Answer: Both acute angles = 45°.
Example 6: Example 6: Complete solution of a triangle
Problem: In △ABC, ∠C = 90°, ∠A = 30°, BC = 6 cm. Find AB and AC.
Solution:
∠B = 60° (since ∠A + ∠B = 90°). BC is opposite to ∠A.
- sin 30° = BC/AB → 1/2 = 6/AB → AB = 12 cm
- cos 30° = AC/AB → √3/2 = AC/12 → AC = 6√3 ≈ 10.39 cm
Verification: BC² + AC² = 36 + 108 = 144 = AB² ✓
Answer: AB = 12 cm, AC = 6√3 cm.
Example 7: Example 7: Tower shadow problem
Problem: A tower casts a shadow of length 20 m when the sun's elevation is 45°. Find the height of the tower.
Solution:
- tan 45° = Height / Shadow
- 1 = Height / 20
- Height = 20 m
Answer: Height = 20 m.
Example 8: Example 8: Staircase problem
Problem: A staircase makes 30° with the floor. If the vertical height is 3.5 m, find the length of the staircase.
Solution:
- sin 30° = 3.5 / Length
- 1/2 = 3.5 / Length
- Length = 7 m
Answer: Staircase length = 7 m.
Example 9: Example 9: Two-angle problem
Problem: A flagpole stands on a building. From a point on the ground 50 m away, the angle of elevation to the top of the building is 30° and to the top of the flagpole is 45°. Find the height of the flagpole.
Solution:
- Height of building: tan 30° = h₁/50 → h₁ = 50/√3 ≈ 28.87 m
- Total height (building + flag): tan 45° = h₂/50 → h₂ = 50 m
- Height of flagpole = 50 − 28.87 = 21.13 m
Answer: Flagpole height ≈ 21.13 m (= 50 − 50/√3 = 50(1 − 1/√3)).
Example 10: Example 10: Isosceles triangle using trigonometry
Problem: An isosceles triangle has equal sides of 10 cm and vertex angle 120°. Find the base and area using trigonometry.
Solution:
The altitude from the vertex to the base bisects the vertex angle (60° each) and bisects the base.
- In the right triangle: hypotenuse = 10, angle at vertex = 60°
- Half-base = 10 × sin 60° = 10 × √3/2 = 5√3
- Base = 10√3 ≈ 17.32 cm
- Height = 10 × cos 60° = 10 × 1/2 = 5 cm
- Area = ½ × base × height = ½ × 10√3 × 5 = 25√3 ≈ 43.3 cm²
Answer: Base = 10√3 cm, Area = 25√3 cm².
Real-World Applications
Applications of trigonometry in right triangles:
- Surveying: Measuring heights of buildings, mountains, and towers.
- Navigation: Finding distances and bearings.
- Construction: Calculating ramp lengths, roof slopes, staircase angles.
- Physics: Resolving forces into horizontal and vertical components.
- Astronomy: Measuring distances to stars using parallax (trigonometric parallax).
- Aviation: Descent angle and glide slope calculations.
Key Points to Remember
- SOH-CAH-TOA is the memory aid for sin, cos, tan definitions.
- In a right triangle, if one acute angle and one side are known, all other parts can be found.
- Memorise exact values for 30°, 45°, 60°.
- The sum of acute angles in a right triangle = 90°. If one is θ, the other is (90° − θ).
- sin θ = cos(90° − θ) and cos θ = sin(90° − θ) — complementary angles.
- Always identify: which side is opposite, adjacent, and hypotenuse relative to the given angle.
- Use Pythagoras theorem to verify or find the third side.
- For height and distance problems: draw a clear figure, mark the angle of elevation/depression, then apply the appropriate ratio.
Practice Problems
- In a right triangle, hypotenuse = 26 cm, one angle = 30°. Find both legs.
- A ladder 10 m long makes 60° with the ground. Find the height reached and the distance from the wall.
- From the top of a 50 m building, the angle of depression to a car is 45°. How far is the car from the base?
- A tree breaks due to wind. The top touches the ground at 30° and 12 m from the base. Find the original height.
- In △PQR, ∠R = 90°, PQ = 15, ∠P = 45°. Find QR and PR.
- A kite string is 100 m long and makes 60° with the ground. Find the height of the kite.
- Two buildings are 50 m apart. From the top of one (20 m tall), the angle of elevation to the top of the other is 30°. Find the height of the taller building.
- Find the area of an equilateral triangle with side 8 cm using sin 60°.
Frequently Asked Questions
Q1. What does 'solving a right triangle' mean?
Finding all unknown sides and angles of a right triangle when sufficient information (an angle and a side, or two sides) is given.
Q2. What is SOH-CAH-TOA?
A mnemonic: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.
Q3. How do I decide which ratio to use?
Identify the given side and the required side relative to the known angle. If you have opposite and hypotenuse, use sin. If adjacent and hypotenuse, use cos. If opposite and adjacent, use tan.
Q4. What if I know two sides but no angle?
Use Pythagoras to find the third side. To find an angle, use the inverse ratio: θ = sin⁻¹(O/H) or θ = tan⁻¹(O/A). At Class 10 level, match with standard angles.
Q5. What is the angle of elevation?
The angle formed between the horizontal line of sight and the line looking upward to an object. Used when looking UP at something.
Q6. What is the angle of depression?
The angle between the horizontal line of sight and the line looking downward to an object. Used when looking DOWN from a height.
Q7. Are angles of elevation and depression equal?
Yes, in the standard setup. The angle of elevation from ground to the top equals the angle of depression from the top to the ground (alternate interior angles with horizontal lines).
Q8. Can trigonometry be used for non-right triangles?
Yes, using the sine rule and cosine rule (covered in Class 11). But in Class 10, we only apply trig to right triangles.
