Trigonometric Ratios of Specific Angles for CBSE Class 10 Mathematics

In general, determining the trigonometric ratios for arbitrary angles requires specialized trigonometric tables or modern scientific calculators. However, for certain angles that appear frequently in geometry, surveying, and architecture—specifically $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$—their trigonometric ratios can be derived geometrically with absolute mathematical precision.

In CBSE Class 10 Mathematics, Chapter 8 (Introduction to Trigonometry) expects students to know these exact algebraic values by heart and be able to evaluate numerical expressions and solve trigonometric equations.

What You Will Learn

• Geometric derivation of trigonometric ratios for $45^\circ$ using an isosceles right triangle
• Geometric derivation of trigonometric ratios for $30^\circ$ and $60^\circ$ using an equilateral triangle
• Geometric understanding of $0^\circ$ and $90^\circ$ as limiting cases
• The Master Values Table across all six trigonometric functions
• The Hand Trick / Fractional Roots memory shortcut
• Trends: How $\sin \theta, \cos \theta,$ and $\tan \theta$ change as $\theta$ increases from $0^\circ$ to $90^\circ$
• Solved CBSE board examination numerical problems and angle-solving techniques

1. Geometric Derivations of Exact Values

1. Trigonometric Ratios of $45^\circ$

Consider an isosceles right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$ and $\angle A = 45^\circ$:

• Since the sum of angles is $180^\circ$, $\angle C = 180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Since $\angle A = \angle C = 45^\circ$, the opposite sides are equal: $BC = AB = a$.
• By the Pythagoras Theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$

Computing Ratios for $45^\circ$:

$\sin 45^\circ = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \mathbf{\frac{1}{\sqrt{2}}}$ $\cos 45^\circ = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \mathbf{\frac{1}{\sqrt{2}}}$ $\tan 45^\circ = \frac{BC}{AB} = \frac{a}{a} = \mathbf{1}$ $\csc 45^\circ = \sqrt{2}, \quad \sec 45^\circ = \sqrt{2}, \quad \cot 45^\circ = 1$

2. Trigonometric Ratios of $30^\circ$ and $60^\circ$

Consider an equilateral triangle $\Delta ABC$ with side length $2a$:

• Each interior angle is $60^\circ$ ($\angle A = \angle B = \angle C = 60^\circ$).
• Draw an altitude $AD \perp BC$.
• In an equilateral triangle, altitude $AD$ bisects base $BC$ and bisects vertex angle $\angle A$: $BD = a, \quad \angle BAD = 30^\circ, \quad \angle B = 60^\circ$
• In right triangle $\Delta ADB$, find $AD$ using Pythagoras theorem: $AD = \sqrt{AB^2 - BD^2} = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \sqrt{3a^2} = a\sqrt{3}$

                                      A (30°)
                                     /|                                     / |                                  2a /  |   \ 2a
                                  /a√3|                                     /    |                               (60°) B-----D------C (60°)
                                   a     a

Computing Ratios for $30^\circ$ (from $\Delta ADB$, with reference angle ngle BAD = 30^\circ):

• $\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \mathbf{\frac{1}{2}}$
• $\cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \mathbf{\frac{\sqrt{3}}{2}}$
• $\tan 30^\circ = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}}$

Computing Ratios for $60^\circ$ (from $\Delta ADB$, with reference angle ngle B = 60^\circ):

• $\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \mathbf{\frac{\sqrt{3}}{2}}$
• $\cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \mathbf{\frac{1}{2}}$
• $\tan 60^\circ = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \mathbf{\sqrt{3}}$

2. The Master Values Table

The Rapid Memory Trick for Sine Values:

Write numbers $0, 1, 2, 3, 4$, divide each by $4$, and take square roots: $\sqrt{\frac{0}{4}} = 0, \quad \sqrt{\frac{1}{4}} = \frac{1}{2}, \quad \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}, \quad \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}, \quad \sqrt{\frac{4}{4}} = 1$ To find cosine, simply reverse the sine sequence from right to left!

3. Behavioral Trends of Ratios ($0^\circ \le heta \le 90^\circ$)

1. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\sin \theta$ increases from $0$ to $1$.
2. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\cos \theta$ decreases from $1$ to $0$.
3. The values of $\sin \theta$ and $\cos \theta$ never exceed $1$ (since hypotenuse is always greater than or equal to perpendicular and base).
4. $\sin \theta = \cos \theta$ occurs at strictly one acute angle: $\theta = 45^\circ$.

4. Solved CBSE Board Examination Problems

Solved Example 1: Numerical Evaluation

Problem: Evaluate: $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.

Solution:

1. Substitute exact values from the table:
   • $\tan 45^\circ = 1$
   • $\cos 30^\circ = \frac{\sqrt{3}}{2}$
   • $\sin 60^\circ = \frac{\sqrt{3}}{2}$
2. Substitute into expression: $2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$ $= 2(1) + \frac{3}{4} - \frac{3}{4} = 2 + 0 = 2$
3. Therefore, <u>the value is $2$</u>.

Solved Example 2: Solving for Unknown Angles (CBSE Classic)

Problem: If $\sin(A - B) = \frac{1}{2}$ and $\cos(A + B) = \frac{1}{2}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the values of $A$ and $B$.

Solution:

1. From $\sin(A - B) = \frac{1}{2}$: Since $\sin 30^\circ = \frac{1}{2}$, we have: $A - B = 30^\circ \quad \text{--- (1)}$
2. From $\cos(A + B) = \frac{1}{2}$: Since $\cos 60^\circ = \frac{1}{2}$, we have: $A + B = 60^\circ \quad \text{--- (2)}$
3. Add Equation (1) and Equation (2): $(A - B) + (A + B) = 30^\circ + 60^\circ$ $2A = 90^\circ \implies A = 45^\circ$
4. Substitute $A = 45^\circ$ into (2): $45^\circ + B = 60^\circ \implies B = 60^\circ - 45^\circ = 15^\circ$
5. Therefore, <u>$A = 45^\circ$ and $B = 15^\circ$</u>.

5. Summary and Examination Tips

Exam Tip: In board exams, take 30 seconds at the start of your rough work sheet to write down the values table. Having the table pre-written prevents high-stress memory blanks during multi-step calculations!

Common Mistake: Writing $\tan 90^\circ = 0$. $\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0}$, which is not defined (undefined), NOT zero!