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Class 9 SELINA Solutions Maths Chapter 22 - Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]

Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise Ex. 22(B)

Solution 1

Consider the given figure

 

(i)

Since the triangle is a right angled triangle, so using Pythagorean Theorem

(ii)

(iii)

Therefore

Solution 2

Consider the given figure

Since the triangle is a right angled triangle, so using Pythagorean Theorem

Also

(i)

(ii)

(iii)

Therefore

Solution 3

Consider the given figure

Since the triangle is a right angled triangle, so using Pythagorean Theorem

In and , the is common to both the triangles, so therefore .

Therefore and are similar triangles according to AAA Rule

So

(i)

(ii)

Solution 4

Consider the given figure

 

Since the triangle is a right angled triangle, so using Pythagorean Theorem

In and , the is common to both the triangles, so therefore.

Therefore and are similar triangles according to AAA Rule

So

Now using Pythagorean Theorem

Therefore


(i) 

 

 

 

(ii)

Solution 5

Consider the figure below

In the isosceles , and the perpendicular drawn from angle to the side divides the side into two equal parts

Solution 6

Consider the figure below

In the isosceles , and the perpendicular drawn from angle to the side divides the side into two equal parts

Since

(i)

(ii)

(iii)

Therefore

(iv)

Therefore

Solution 7

Consider the figure


Therefore if length of base = 4x, length of perpendicular = 3x


Since


Now


Therefore


And

Solution 8

Consider the figure

 

A perpendicular is drawn from D to the side AB at point E which makes BCDE is a rectangle.

Now in right angled triangle BCD using Pythagorean Theorem

Since BCDE is rectangle so ED 12 cm, EB = 5 and AE = 14 - 5 = 9 

(i)

(ii)

Or

Solution 9

Given



Therefore if length of perpendicular = 4x, length of hypotenuse = 5x


Since

Now

(i)

And

(ii)

Given


Therefore if length of perpendicular = x, length of hypotenuse = x


Since

Now

So

And


Now


So

Solution 10

Squaring both sides

Solution 11

Squaring both sides

Solution 12

Consider the diagram below:


Therefore if length of BC = 3x, length of AB = 4x


Since

(i)

(ii)

(iii)

Therefore

(iv)

Solution 13

Consider the diagram below:


Therefore if length of AB = 15x, length of AC = 17x


Since


Now


Therefore

Solution 14

Now

Solution 15

Since is mid-point of so

(i)

(ii)

Solution 16

Consider the diagram below:


Therefore if length of AB = 4x, length of BC = 3x


Since


(i)


(ii)


Therefore

Solution 17

Consider the figure


Therefore if length of base = 4x, length of perpendicular = 3x


Since


Now


Therefore


And

Solution 18

Consider the figure

The diagonals of a rhombus bisects each other perpendicularly


Therefore if length of base = 3x, length of hypotenuse = 5x


Since


Now


Therefore

And


Since the sides of a rhombus are equal so the length of the side of the rhombus


The diagonals are

Solution 19

Consider the figure below

In the isosceles , the perpendicular drawn from angle to the side divides the side into two equal parts

Since

(i)

(ii)

(iii)

Therefore

Solution 20

Consider the figure below


Therefore if length of perpendicular = 4x, length of hypotenuse = 5x


Since


Now


Therefore


And

Again


Therefore if length of perpendicular = x, length of base = x


Since


Now


Therefore


And

Solution 21

Now

Solution 22

Consider the figure


Therefore if length of perpendicular = x, length of base = x


Since


Now


Therefore

Solution 23

Consider the diagram


Therefore if length of base = 3x, length of perpendicular = 2x


Since


Now


Therefore


Now

Solution 24

Consider the figure

Therefore if length of , length of

Since

Now

(i)

(ii)

Solution 25

Consider the diagram below:

Therefore if length of , length of

Since

Consider the diagram below:

Therefore if length of , length of

Since

Now

Therefore

Solution 26

Consider the given diagram as

Using Pythagorean Theorem

Now

Again using Pythagorean Theorem

Now

Therefore

Solution 27

Consider the figure

Therefore if length of , length of

Since

Now

Therefore

Solution 28

Now

Solution 29

Squaring both sides

Solution 30

Now

(i)

(ii)

Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise Ex. 22(A)

Solution 1

Given angle

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution 2

Given angle

(i)

(ii)

(iii)

(iv)

Solution 3

Consider the diagram as

 

Given angle and

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution 4

Given angle and

(i)

(ii)

(iii)

(iv)

Solution 5

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

Solution 6

Given angle in the figure

Now

(i)

(ii)

(iii)

 

Solution 7

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

Solution 8

Consider the diagram below:

Therefore if length of , length of

Since

Now

Therefore

Solution 9

Consider the diagram below:


Therefore if length of A B equals 3 x, length of B C equals 4 x

Since


Now


Therefore

Solution 10

Consider the diagram below:

 


Therefore if length of AB = 3x, length of BC = 4x

Since

(i)

(ii)

(iii)

Solution 11

Consider the diagram below:


Therefore if length of AB = 3x, length of AC = 5x

Since


Now all other trigonometric ratios are

Solution 12

Consider the diagram below:


Therefore if length of AB = 12x, length of BC = 5x

Since


(i)

(ii)

(iii)

Solution 13

Consider the diagram below:


Therefore if length of perpendicular = px, length of hypotenuse = qx


Since


Now


Therefore

Solution 14

Consider the diagram below:


Therefore if length of AB = x, length of AC = 2x

Since


Consider the diagram below:

 


Therefore if length of AC = x, length of B C equals square root of 2 x 

Since


Now


Therefore

Solution 15

Consider the diagram below:


Therefore if length of base = 12x, length of perpendicular = 5x


Since


Now


Therefore

Solution 16

Consider the diagram below:


Therefore if length of base = 3x, length of perpendicular = 4x


Since


Now


Therefore

Solution 17

Consider the diagram below:

 


Therefore if length of hypotenuse equals square root of 5 x, length of perpendicular = x

Since


Now

(i)

 

(ii)

Solution 18

Consider the diagram below:


Therefore if length of AB = x, length of A C equals square root of 2 x 

Since


Now


Therefore

Solution 19

Consider the diagram below:


Therefore if length of base = x, length of perpendicular = x


Since


Now


Therefore

Solution 20

Given angle and in the figure


Again


Now

(i)

(ii)


Therefore


(iii)


Therefore

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