RD sharma class 10 Chapter 8 Trigonometric Identities

Chapter 3 Linear Equations in Two Variables Download Pdf of RD sharma class 10 Chapter 8 Trigonometric Identities

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RD sharma class 10 Chapter 5 Trigonometric Ratios

Chapter 3 Linear Equations in Two Variables Download Pdf of RD sharma class 10 Chapter 5 Trigonometric Ratios

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Class 10 Chapter 9 (some applications on trigonometry) class notes

Notes on Some Applications of Trigonometry

Trigonometry can be used in many ways in the things around us like we can use it for calculating the height and distance of some objects without calculating them actually.

Heights and Distances

To find the height of an object or to find the distance of an object to the other we must know the meaning of some points -

Line of Sight

When we look at some object then the line made by our vision to the object is called the Line of Sight.

Horizontal Line

A horizontal line is a distance between the observer and the object.

Angle of Elevation

An angle of elevation is the angle made by the line of sight to the top of the object and the horizontal line. It is above the horizontal line i.e. the angle of elevation is made when we look up to the object.

Angle of Depression

An angle of depression is made when the observer needs to look down to see the object.The angle between the horizontal line and the line of sight is the angle of depression when the horizontal line is above the angle.

Some Solved Examples

To solve the problems related to the angle of elevation and angle of depression we must remember trigonometric ratios, trigonometric table and the trigonometric identities.

Example: 1

Find the height of the flagpole if the angle of elevation is 30° and the distance of flag from the observer is 15 m.

Solution:

Let x be the height of the flagpole.

Horizontal line is 15m.

Now, to calculate x, we need to take the trigonometric ratio which includes perpendicular and base.

Hence, the flagpole is 9 m.

Example: 2

A child was playing at the top of the hill. He had thrown a stone in the lake from the hill, the distance covered by the stone was 150 m and the angle of depression was 30°, then what is the height of the hill?

Solution:

Let the height of the hill be h.

The hypotenuse is 150 m and the angle of depression is 30°.

Hence the height of the hill is 75 m.

Example: 3

A person standing at point A is looking up at the angle of elevation of 45° to the aeroplane which is at the height of 100 m. As the airplane is going upwards, after some, the person was looking at the angle of elevation of 60°. Then what will be the increase in the height of the aeroplane from the ground at the angle of 60°?

Solution:

Given

∠CAB = 45°, ∠DAB = 60°

Distance of the aero plane from the ground = x + 100 m

In ∆ABC

AB = BC = 100 m

In ∆ABD

BD = BC + CD

100√3m = 100 + x

x = 100(√3 - 1) m

Hence, the increase in the height of the aero plane is100 (√3 -1) m.

Example: 4

A girl who is 1.2 m tall is watching a ballon moving in a horizontal line at a height of 88.2 m from the ground.The angle of elevation from her eyes is changed from 60° to 30°. Calculate the distance travelled by balloon.

Solution:

In ΔACE

In ΔBCG

CG = 87√3 m

Distance travelled by ballon = EG = CG - CE

87√3 - 29√3 = 58√3 m

cbse class 10th Heights and distances

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Class 10 Chapter 8 (Introduction to Trigonometry) class notes

Notes on Introduction to Trigonometry

Trigonometry

To find the distances and heights we can use the mathematical techniques, which come under the Trigonometry. It shows the relationship between the sides and the angles of the triangle. Generally, it is used in the case of a right angle triangle.

Trigonometric Ratios

In a right angle triangle, the ratio of its side and the acute angles is the trigonometric ratios of the angles.

In this right angle triangle $\angle B={90}^{\circ }$. If we take $\angle A$ as acute angle then -

AB is the base, as the side adjacent to the acute angle.

BC is the perpendicular, as the side opposite to the acute angle.

Ac is the hypotenuse, as the side opposite to the right angle.

Trigonometric ratios with respect to ∠A

Ratio	Formula	Short form	Value
sin A		$\frac{P}{H}$	$\frac{BC}{AC}$
cos A		$\frac{B}{H}$	$\frac{AB}{AC}$
tan A		$\frac{P}{B}$	$\frac{BC}{AB}$
cosec A		$\frac{H}{P}$	$\frac{AC}{BC}$
sec A		$\frac{H}{B}$	$\frac{AC}{AB}$
cot A		$\frac{B}{P}$	$\frac{AB}{BC}$

Remark

• If we take ∠C as acute angle then BC will be base and AB will be perpendicular. Hypotenuse remains the same i.e. AC.So the ratios will be according to that only.
• If the angle is same then the value of the trigonometric ratios of the angles remain the same whether the length of the side increases or decreases.
• In a right angle triangle, the hypotenuse is the longest side so sin A or cos A will always be less than or equal to 1 and the value of sec A or cosec A will always be greater than or equal to 1.

Reciprocal Relation between Trigonometric Ratios

Cosec A, sec A, and cot A are the reciprocals of sin A, cos A, and tan A respectively.

Quotient Relation

Trigonometric Ratios of Some Specific Angles

Use of Trigonometric Ratios and Table in Solving Problems

Example

Find the lengths of the sides BC and AC in ∆ ABC, right-angled at B where AB = 25 cm and ∠ACB = 30°, using trigonometric ratios.

Solution

To find the length of the side BC, we need to choose the ratio having BC and the given side AB. As we can see that BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore

tan 30°

BC = 25√3 cm

To find the length of the side AC, we consider

AC = 50 cm

Trigonometric Ratios of Complementary Angles

If the sum of two angles is 90° then, it is called Complementary Angles. In a right-angled triangle, one angle is 90°, so the sum of the other two angles is also 90° or they are complementary angles.so the trigonometric ratios of the complementary angles will be -

sin (90° – A) = cos A,

cos (90° – A) = sin A,

tan (90° – A) = cot A,

cot (90° – A) = tan A,

sec (90° – A) = cosec A,

cosec (90° – A) = sec A

Trigonometric Identities (Pythagoras Identity)

An equation is said to be a trigonometric identity if it contains trigonometric ratios of an angle and satisfies it for all values of the given trigonometric ratios.

In ∆PQR, right angled at Q, we can say that

PQ² + QR² = PR²

Divide each term by PR², we get

$\begin{array}{l}\frac{P{Q}^2}{P{R}^2}+\frac{Q{R}^2}{P{R}^2}=\frac{P{R}^2}{P{R}^2}\\ {\left(\frac{PQ}{PR}\right)}^2+{\left(\frac{QR}{PR}\right)}^2=1\\ {\sin }^{2}R+{\cos }^{2}R=1\end{array}$

Likewise other trigonometric identities can also be proved. So the identities are-

$\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ 1+{\tan }^2\theta ={\sec }^2\theta \\ {\cot }^2\theta +1=\cos e{c}^2\theta \end{array}$

How to solve the problems related to trigonometric ratios and identities?

Prove that

cbse class 10 introduction to trigonometry

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