Trigonometric Ratios of an Acute Angle for CBSE Class 10 Mathematics

Imagine standing on the ground looking up at the summit of a towering mountain or the spire of a historic temple. How could ancient surveyors calculate the exact height of such monuments without physically climbing them with a measuring tape? The answer lies in trigonometry—a Greek word combining tri (three), gon (sides), and metron (measure)—the study of relationships between the sides and angles of a triangle.

In CBSE Class 10 Mathematics, Chapter 8 (Introduction to Trigonometry) introduces the six fundamental trigonometric ratios. These ratios form the universal language connecting linear distances to rotational angles across engineering, physics, and advanced mathematics.

What You Will Learn

• Right-angled triangle anatomy: Hypotenuse, Perpendicular (Opposite), and Base (Adjacent)
• Why the reference angle dictates which side is Perpendicular and which is Base
• The six trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent
• Reciprocal relations and quotient relations
• Why trigonometric ratios depend solely on the angle, not the size of the triangle
• Step-by-step methods to deduce all six ratios when one ratio is given
• Board exam questions, presentation templates, and common student errors

1. Anatomy of a Right-Angled Triangle

Consider a right-angled triangle $\Delta ABC$, right-angled at vertex $B$ ($\angle B = 90^\circ$). Let us inspect the sides relative to an acute angle $\angle A$ (or $\theta$):

                       A
                       |                       |        Base (Adjacent) |  \  Hypotenuse (Longest side)
       to angle A      |                          |                           +-----+
                       B     C
                    Perpendicular (Opposite to angle A)

1. Hypotenuse ($H$): The side opposite the $90^\circ$ right angle ($AC$). It is always the longest side of the right triangle.
2. Perpendicular / Opposite Side ($P$): The side directly opposite to the reference angle $\angle A$ (here, side $BC$).
3. Base / Adjacent Side ($B$): The side adjacent to the reference angle $\angle A$ (here, side $AB$).

Important: <u>The designations "Perpendicular" and "Base" are NOT fixed; they depend strictly on which acute angle is being observed! If you observe from $\angle A$, side $BC$ is the Perpendicular and $AB$ is the Base. But if you observe from $\angle C$, side $AB$ becomes the Perpendicular and $BC$ becomes the Base!</u>

2. The Six Fundamental Trigonometric Ratios

For an acute angle $\theta$ in a right-angled triangle:

1. Primary Trigonometric Ratios:

• Sine of $\theta$ ($\sin \theta$): $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{P}{H}$
• Cosine of $\theta$ ($\cos \theta$): $\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{B}{H}$
• Tangent of $\theta$ ($\tan \theta$): $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{P}{B}$

2. Reciprocal Trigonometric Ratios:

• Cosecant of $\theta$ ($\csc \theta$ or $\text{cosec } \theta$): $\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{H}{P}$
• Secant of $\theta$ ($\sec \theta$): $\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{H}{B}$
• Cotangent of $\theta$ ($\cot \theta$): $\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Base}}{\text{Perpendicular}} = \frac{B}{P}$

The Famous Mnemonic:

To memorize the primary ratios easily, remember: $\mathbf{\text{Some People Have, Curly Brown Hair, Turned Permanent Black}}$ $\sin = \frac{P}{H}, \quad \cos = \frac{B}{H}, \quad \tan = \frac{P}{B}$

3. Quotient Relations

Dividing $\sin \theta$ by $\cos \theta$: $\frac{\sin \theta}{\cos \theta} = \frac{P/H}{B/H} = \frac{P}{B} = \mathbf{\tan \theta}$

Similarly, dividing $\cos \theta$ by $\sin \theta$: $\frac{\cos \theta}{\sin \theta} = \frac{B/H}{P/H} = \frac{B}{P} = \mathbf{\cot \theta}$

Remember: $\sin A$ is an abbreviation for "the sine of angle $A$". It is NOT the product of $\sin$ and $A$. $\sin$ separated from an angle has no mathematical meaning!

4. Invariance of Ratios with Triangle Size

Does the value of $\sin \theta$ change if you make the triangle larger?

• Consider two right-angled triangles of different sizes sharing the same acute angle $\theta$.
• By AA similarity, the two triangles are similar.
• Since corresponding sides of similar triangles are in the exact same proportion, the ratios $\frac{P}{H}, \frac{B}{H},$ and $\frac{P}{B}$ remain completely identical regardless of the size of the triangle!

5. Solved CBSE Board Examination Problems

Solved Example 1: Finding All Ratios from One Given Ratio

Problem: Given $\tan A = \frac{4}{3}$, find the other trigonometric ratios of the angle $A$.

Solution:

1. Represent the situation geometrically: Consider a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$. We know that: $\tan A = \frac{\text{Side opposite to } \angle A}{\text{Side adjacent to } \angle A} = \frac{BC}{AB} = \frac{4}{3}$
2. Assign a positive scaling constant $k$: Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive real number.
3. Find the hypotenuse $AC$ using Pythagoras Theorem: $AC^2 = AB^2 + BC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$ $AC = \sqrt{25k^2} = 5k$
4. Compute the remaining five trigonometric ratios:
   • $\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \mathbf{\frac{4}{5}}$
   • $\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \mathbf{\frac{3}{5}}$
   • $\csc A = \frac{1}{\sin A} = \mathbf{\frac{5}{4}}$
   • $\sec A = \frac{1}{\cos A} = \mathbf{\frac{5}{3}}$
   • $\cot A = \frac{1}{\tan A} = \mathbf{\frac{3}{4}}$

Solved Example 2: Evaluating Algebraic Trigonometric Expressions

Problem: In $\Delta PQR$, right-angled at $Q$, $PR + QR = 25\text{ cm}$ and $PQ = 5\text{ cm}$. Determine the values of $\sin P$, $\cos P$, and $\tan P$.

Solution:

1. Let $QR = x\text{ cm}$. Then $PR = (25 - x)\text{ cm}$. We are given $PQ = 5\text{ cm}$.
2. In right triangle $\Delta PQR$, by Pythagoras Theorem: $PR^2 = PQ^2 + QR^2$ $(25 - x)^2 = 5^2 + x^2$
3. Expand LHS: $625 - 50x + x^2 = 25 + x^2$
4. Cancel $x^2$ on both sides: $625 - 50x = 25 \implies 50x = 600 \implies x = 12\text{ cm}$
5. Side lengths:
   • $QR = x = 12\text{ cm}$ (Perpendicular relative to $\angle P$)
   • $PR = 25 - 12 = 13\text{ cm}$ (Hypotenuse)
   • $PQ = 5\text{ cm}$ (Base relative to $\angle P$)
6. Evaluate ratios for $\angle P$: $\sin P = \frac{QR}{PR} = \mathbf{\frac{12}{13}}$ $\cos P = \frac{PQ}{PR} = \mathbf{\frac{5}{13}}$ $\tan P = \frac{QR}{PQ} = \mathbf{\frac{12}{5}}$

6. Summary and Examination Tips

Exam Tip: When given a ratio like $\tan A = \frac{4}{3}$, never write $BC = 4$ and $AB = 3$ directly without declaring a positive constant $k$ ($BC = 4k, AB = 3k$). Writing without $k$ can result in the deduction of half a mark in board exams!

Common Mistake: Writing $(\sin \theta)^2$ as $\sin \theta^2$. The square of the sine ratio is written as $\sin^2 \theta$. Writing $\sin \theta^2$ means the sine of the angle squared!