Trigonometric Ratios, Heights and Distances
Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. It is particularly useful in solving problems involving indirect measurements, such as determining the height of a tall building or the distance across a river. The core of trigonometry lies in trigonometric ratios, which are defined for a right-angled triangle.
Trigonometric Ratios
Consider a right-angled triangle ABC, where the angle at B is 90 degrees. Let A be one of the acute angles.
- Hypotenuse (H): The side opposite the right angle (AC).
- Perpendicular (P) / Opposite Side: The side opposite to the angle A (BC).
- Base (B) / Adjacent Side: The side adjacent to the angle A (AB).
The six fundamental trigonometric ratios are defined as follows, with respect to angle A:
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Sine (sin A): The ratio of the length of the perpendicular to the length of the hypotenuse.
sin A = Perpendicular / Hypotenuse = P / H
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Cosine (cos A): The ratio of the length of the base to the length of the hypotenuse.
cos A = Base / Hypotenuse = B / H
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Tangent (tan A): The ratio of the length of the perpendicular to the length of the base.
tan A = Perpendicular / Base = P / B
It's also important to note that tan A = sin A / cos A.
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Cosecant (csc A or cosec A): The reciprocal of sine A.
csc A = Hypotenuse / Perpendicular = H / P = 1 / sin A
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Secant (sec A): The reciprocal of cosine A.
sec A = Hypotenuse / Base = H / B = 1 / cos A
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Cotangent (cot A): The reciprocal of tangent A.
cot A = Base / Perpendicular = B / P = 1 / tan A
Also, cot A = cos A / sin A.
P/H = sin A, B/H = cos A, P/B = tan A.
The other three are their reciprocals: H/P = cosec A, H/B = sec A, B/P = cot A.
Standard Angles and Their Trigonometric Ratios
Certain angles have well-defined and commonly used trigonometric ratios. These are essential for solving problems quickly.
| Angle (θ) | sin θ | cos θ | tan θ | csc θ | sec θ | cot θ |
|---|---|---|---|---|---|---|
| 0° | 0 | 1 | 0 | Undefined | 1 | Undefined |
| 30° | 1/2 | √3/2 | 1/√3 | 2 | 2/√3 | √3 |
| 45° | 1/√2 | 1/√2 | 1 | √2 | √2 | 1 |
| 60° | √3/2 | 1/2 | √3 | 2/√3 | 2 | 1/√3 |
| 90° | 1 | 0 | Undefined | 1 | Undefined | 0 |
For sin values: Write 0, 1, 2, 3, 4. Divide each by 4: 0/4, 1/4, 2/4, 3/4, 4/4. Take the square root: 0, 1/2, 1/√2, √3/2, 1. This gives sin 0°, 30°, 45°, 60°, 90°.
For cos values: Write the sin values in reverse order.
For tan values: tan θ = sin θ / cos θ. Divide the sin value by the corresponding cos value.
For cosec, sec, cot: Take the reciprocals of sin, cos, and tan values respectively.
Trigonometric Identities
Identities are equations that are true for all values of the variables involved. The most fundamental identities are derived from the Pythagorean theorem. In a right-angled triangle with sides P, B, and H, we know that P² + B² = H². Dividing this by H² gives:
(P²/H²) + (B²/H²) = H²/H²
(sin A)² + (cos A)² = 1
This leads to the fundamental Pythagorean identity:
sin² A + cos² A = 1
From this, we can derive other important identities:
- 1 + tan² A = sec² A
- 1 + cot² A = csc² A
Other useful identities include:
- sin (90° - A) = cos A
- cos (90° - A) = sin A
- tan (90° - A) = cot A
- csc (90° - A) = sec A
- sec (90° - A) = csc A
- cot (90° - A) = tan A
Heights and Distances
This topic involves applying trigonometric ratios to solve real-world problems where direct measurement is difficult. The key concepts are the angle of elevation and the angle of depression.
Angle of Elevation
When an observer looks upwards from a point to an object that is higher than the observer's eye level, the angle formed between the horizontal line from the observer's eye and the line of sight to the object is called the angle of elevation.
Angle of Depression
When an observer looks downwards from a point to an object that is lower than the observer's eye level, the angle formed between the horizontal line from the observer's eye and the line of sight to the object is called the angle of depression.
An important principle here is that the angle of elevation from point A to point B is equal to the angle of depression from point B to point A, because they form alternate interior angles between parallel horizontal lines.
- Read the problem carefully: Understand what is given and what needs to be found.
- Draw a diagram: Sketch a right-angled triangle (or multiple triangles) representing the situation. Label the known angles, sides, and the unknown quantity.
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Identify the trigonometric ratio: Based on the angle and the sides involved (perpendicular, base, hypotenuse), choose the appropriate trigonometric ratio (sin, cos, tan).
- If you have the angle and the adjacent side (base) and want to find the opposite side (height), use tan.
- If you have the angle and the opposite side (height) and want to find the adjacent side (distance), use tan.
- If you have the angle and the opposite side (height) and want to find the hypotenuse (line of sight), use sin.
- If you have the angle and the hypotenuse and want to find the opposite side (height), use sin.
- If you have the angle and the adjacent side (base) and want to find the hypotenuse, use cos.
- If you have the angle and the hypotenuse and want to find the adjacent side (base), use cos.
- Set up the equation: Write the trigonometric equation using the chosen ratio.
- Solve for the unknown: Solve the equation to find the required height or distance. Use the values of trigonometric ratios for standard angles if applicable.
Example Problems
Example 1:
A ladder 10 meters long reaches a window 8 meters above the ground. What is the angle of elevation of the ladder with the horizontal?
Diagram:
Let the ladder be AC, the height of the window from the ground be BC, and the distance of the foot of the ladder from the wall be AB.
Here, AC (hypotenuse) = 10 m, BC (perpendicular) = 8 m. We need to find the angle ACB, let's call it θ.
Solution:
In right-angled triangle ABC, we have the perpendicular and the hypotenuse. The trigonometric ratio relating these is sine.
sin θ = Perpendicular / Hypotenuse = BC / AC = 8 / 10 = 0.8
θ = sin⁻¹(0.8)
Using a calculator, θ ≈ 53.13°.
Note: For competitive exams, angles are usually standard ones like 30°, 45°, 60°, or angles where the sine, cosine, or tangent values are easily derived. If sin θ = 0.8, and it's not a standard angle, the question might ask for sin θ or provide options.
Example 2:
The angle of elevation of the top of a tower from a point on the ground, which is 50 meters away from the foot of the tower, is 60°. Find the height of the tower.
Diagram:
Let the tower be AB, and the point on the ground be C. The distance AC = 50 m. The angle of elevation of the top of the tower (B) from C is ∠BCA = 60°. We need to find the height of the tower AB.
Solution:
In right-angled triangle ABC, we have the base (AC) and we need to find the perpendicular (AB). The angle is 60°. The trigonometric ratio is tangent.
tan 60° = Perpendicular / Base = AB / AC
We know tan 60° = √3.
√3 = AB / 50
AB = 50√3 meters.
Example 3:
From the top of a cliff 100 meters high, the angle of depression of a boat at sea is 30°. Find the distance of the boat from the foot of the cliff.
Diagram:
Let the cliff be AB, with the top at A and the foot at B. The height AB = 100 m. Let the boat be at point C. The angle of depression from A to C is 30°. We need to find the distance BC.
Draw a horizontal line AD from A. The angle of depression ∠DAC = 30°. Since AD is parallel to BC, the alternate interior angle ∠ACB = ∠DAC = 30°.
Solution:
In right-angled triangle ABC, we have the perpendicular (AB) and we need to find the base (BC). The angle is 30°. The trigonometric ratio is tangent.
tan 30° = Perpendicular / Base = AB / BC
We know tan 30° = 1/√3.
1/√3 = 100 / BC
BC = 100√3 meters.
- Always draw a clear diagram.
- Understand the difference between angle of elevation and angle of depression.
- Remember that angle of elevation from A to B equals angle of depression from B to A.
- Use the appropriate trigonometric ratio (SOH CAH TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent).
- For competitive exams, be very familiar with the trigonometric ratios of 30°, 45°, and 60°.
- If multiple angles or heights are involved, you might need to use two right-angled triangles and solve them simultaneously.
Advanced Concepts and Common Pitfalls
While the basics are straightforward, problems can become complex. Be aware of:
- Angles of Elevation/Depression from different points: Sometimes, you might be given information from two different points on the ground, requiring you to set up two triangles and use algebraic equations to solve for unknowns. For example, if the angle of elevation from point A is α and from point B (closer to the object) is β, and the distance between A and B is 'd', you can form two equations.
- Objects at different heights: Problems might involve a person standing on top of a building looking at another object, or two objects at different heights relative to each other. Ensure your diagram correctly represents the horizontal lines and the angles.
- Using reciprocal ratios: While tan is most common, don't shy away from using cot, sec, or csc if they simplify the equation based on the sides you have and need. For instance, if you have the opposite and adjacent sides, cot θ = Adjacent/Opposite might be easier to work with than tan θ = Opposite/Adjacent if the numbers are simpler.
- Units: Always ensure consistency in units (meters, feet, etc.) and clearly state the units in your final answer.
- Approximations: If standard angles are not involved and approximations are needed (e.g., using √3 ≈ 1.732), ensure you follow the instructions in the question regarding the level of precision.
Mastering trigonometric ratios and their application in heights and distances is crucial for the Mathematical Abilities section of the SSC CGL exam. Consistent practice with a variety of problems, focusing on diagrammatic representation and the correct application of ratios, will build confidence and speed.