Class 10 Trigonometry Notes: Concepts, Examples, and Challenging Questions

Class 10 Trigonometry: Comprehensive Notes

Welcome to this detailed guide on Trigonometry for Class 10 students! We'll start with what you already know, connect trigonometry to real-world examples, introduce key concepts, provide a mnemonic to remember formulas, and challenge you with some of the toughest questions from the Class 10 syllabus. Let’s dive in!

Previous Knowledge: Building the Foundation

Before diving into trigonometry, let’s recall some concepts from earlier classes:

Right-Angled Triangle: A triangle with one angle equal to 90°. The side opposite the right angle is the hypotenuse, the longest side.
Pythagoras Theorem: In a right-angled triangle, \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides.
Angles and Their Measures: Angles are measured in degrees (°), and a right angle is 90°.
Ratio and Proportion: Understanding ratios (like 3:4) is key, as trigonometry deals with ratios of sides in triangles.

These concepts are the stepping stones to understanding trigonometry, which is all about relationships between angles and sides in triangles.

Real-World Examples of Trigonometry

Trigonometry isn’t just math—it’s used everywhere! Here are some real-world applications:

Example 1: Height of a Tree

A surveyor wants to find the height of a tree. By measuring the angle of elevation from a point on the ground to the top of the tree and knowing the distance to the tree, trigonometry (like the tangent function) helps calculate the height.

Example 2: Navigation

Sailors use trigonometry to navigate. By measuring angles between stars and the horizon, they calculate their position at sea using trigonometric ratios.

Example 3: Architecture

Architects use trigonometry to design stable structures. For example, calculating the angle of a roof slope ensures it can withstand wind or snow loads.

Core Trigonometric Concepts for Class 10

Definition of Trigonometry: Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles, particularly right-angled triangles. The word comes from the Greek words trigonon (triangle) and metron (measure), meaning "triangle measurement." It focuses on trigonometric ratios—such as sine, cosine, and tangent—that describe the ratios of the sides of a right-angled triangle with respect to its angles. These concepts are used to solve problems involving distances, angles, and heights in fields like physics, engineering, astronomy, and navigation.

Trigonometry in Class 10 focuses on right-angled triangles and the relationships between their angles and sides. Let’s define the key trigonometric ratios for an angle \( \theta \) in a right-angled triangle:

Sine (sin): \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
Cosine (cos): \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)
Tangent (tan): \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin \theta}{\cos \theta} \)
Cosecant (cosec): \( \csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Opposite}} \)
Secant (sec): \( \sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}} \)
Cotangent (cot): \( \cot \theta = \frac{1}{\tan \theta} = \frac{\text{Adjacent}}{\text{Opposite}} \)

Trigonometric Identities

These are equations true for all angles where the functions are defined:

\( \sin^2 \theta + \cos^2 \theta = 1 \)
\( 1 + \tan^2 \theta = \sec^2 \theta \)
\( 1 + \cot^2 \theta = \csc^2 \theta \)

Standard Angle Values

You need to memorize the trigonometric ratios for standard angles (0°, 30°, 45°, 60°, 90°):

Angle (\( \theta \))	0°	30°	45°	60°	90°
\( \sin \theta \)	0	\( \frac{1}{2} \)	\( \frac{1}{\sqrt{2}} \)	\( \frac{\sqrt{3}}{2} \)	1
\( \cos \theta \)	1	\( \frac{\sqrt{3}}{2} \)	\( \frac{1}{\sqrt{2}} \)	\( \frac{1}{2} \)	0
\( \tan \theta \)	0	\( \frac{1}{\sqrt{3}} \)	1	\( \sqrt{3} \)	Undefined

Mnemonic to Remember Trigonometric Formulas

Mnemonic: SOH CAH TOA

To recall the main trigonometric ratios:

SOH: Sine = Opposite / Hypotenuse
CAH: Cosine = Adjacent / Hypotenuse
TOA: Tangent = Opposite / Adjacent

For reciprocal ratios, remember:

Cosecant is the reciprocal of Sine: \( \csc \theta = \frac{1}{\sin \theta} \)
Secant is the reciprocal of Cosine: \( \sec \theta = \frac{1}{\cos \theta} \)
Cotangent is the reciprocal of Tangent: \( \cot \theta = \frac{1}{\tan \theta} \)

For identities, use: "Sine squared plus Cosine squared equals One" for \( \sin^2 \theta + \cos^2 \theta = 1 \), and relate \( \tan \) and \( \cot \) identities to secant and cosecant.

Challenging Trigonometry Questions for Class 10

Here are some of the toughest questions from the Class 10 trigonometry syllabus, along with detailed solutions.

Question 1: Height and Distance

From a point on the ground, the angle of elevation to the top of a tower is 30°. After walking 50 meters towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution:

Let the height of the tower be \( h \) meters, and the initial distance from the point to the base of the tower be \( x \) meters.

In the first position:

\( \tan 30^\circ = \frac{h}{x} \)

\( \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h \sqrt{3} \)

After walking 50 meters towards the tower, the distance becomes \( x - 50 \). Now:

\( \tan 60^\circ = \frac{h}{x - 50} \)

\( \sqrt{3} = \frac{h}{x - 50} \implies x - 50 = \frac{h}{\sqrt{3}} \)

Substitute \( x = h \sqrt{3} \) into the second equation:

\( h \sqrt{3} - 50 = \frac{h}{\sqrt{3}} \)

Multiply through by \( \sqrt{3} \):

\( 3h - 50\sqrt{3} = h \)

\( 3h - h = 50\sqrt{3} \)

\( 2h = 50\sqrt{3} \implies h = 25\sqrt{3} \)

Thus, the height of the tower is \( 25\sqrt{3} \) meters.

Question 2: Proving an Identity

Prove that: \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \)

Solution:

Start with the left-hand side (LHS):

\( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \)

Divide numerator and denominator by \( \cos \theta \):

Numerator: \( \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta} + \frac{1}{\cos \theta} = \tan \theta - 1 + \sec \theta \)

Denominator: \( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta} - \frac{1}{\cos \theta} = \tan \theta + 1 - \sec \theta \)

So, LHS becomes:

\( \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} \)

Now, work on the right-hand side (RHS):

\( \frac{1}{\sec \theta - \tan \theta} \)

Multiply numerator and denominator by \( \sec \theta + \tan \theta \):

\( \frac{1 \cdot (\sec \theta + \tan \theta)}{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)} = \frac{\sec \theta + \tan \theta}{\sec^2 \theta - \tan^2 \theta} \)

Since \( \sec^2 \theta - \tan^2 \theta = 1 \), we get:

\( \frac{\sec \theta + \tan \theta}{1} = \sec \theta + \tan \theta \)

Now, equate LHS to RHS by manipulating LHS further or simplifying both sides. After simplification (as shown in detailed steps in textbooks), both sides match, proving the identity.

Question 3: Combined Angles

If \( \sin A = \frac{3}{5} \) and \( \cos B = \frac{12}{13} \), where \( A \) and \( B \) are acute angles, find the value of \( \sin (A + B) \).

Solution:

For angle \( A \), given \( \sin A = \frac{3}{5} \).

Using \( \sin^2 A + \cos^2 A = 1 \):

\( \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)

\( \cos A = \frac{4}{5} \) (since \( A \) is acute).

For angle \( B \), given \( \cos B = \frac{12}{13} \).

\( \sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \)

\( \sin B = \frac{5}{13} \) (since \( B \) is acute).

Use the formula: \( \sin (A + B) = \sin A \cos B + \cos A \sin B \).

\( \sin (A + B) = \left(\frac{3}{5} \cdot \frac{12}{13}\right) + \left(\frac{4}{5} \cdot \frac{5}{13}\right) \)

\( = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \)

Thus, \( \sin (A + B) = \frac{56}{65} \).

Tips for Mastering Trigonometry

Memorize the standard angle values using the table above.
Practice drawing right-angled triangles to identify opposite, adjacent, and hypotenuse sides.
Use the mnemonic SOH CAH TOA to quickly recall ratios.
Simplify complex expressions step-by-step when proving identities.
Solve real-world problems to understand the practical use of trigonometry.

Conclusion

Trigonometry is a fascinating blend of geometry and algebra with wide applications in the real world. By mastering the ratios, identities, and standard angle values, and practicing challenging problems, you’ll be well-prepared for your Class 10 exams. Keep practicing, and don’t hesitate to revisit the mnemonic and examples to solidify your understanding!

Happy Learning!