NCERT CLASS X Trigonometry

 Simple Explanation of trigonometry for class X

Overview of Trigonometry
• A field of mathematics that focuses on particular angles' functions.
• The sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc) are the six common functions.

Before computers rendered trigonometry tables unnecessary, calculated values were tabulated for many angles. Functions are properties of the angle A, regardless of the size of the triangle. In geometric figures, unknown angles and distances can be calculated using trigonometric functions from known or measured angles.

CosecӨ = 1/ SinӨ 

Sec Ө = 1/Cos Ө

Cot Ө = 1/tan Ө

=> So, we can see

SinӨ * Cosec Ө = 1

CosӨ * SecӨ = 1

Tan(θ)*Cot(θ) = 1

* We will use Pythagoras' theorem multiple times to evaluate any of the missing values in the right-angled triangle. 

Exercise 8

Question 1 

Given that in the triangle ABC, AB = 24, BC = 7 

Using Pythagoras' theorem

AC² = AB² + BC²

AC²= 24²+ 7²

        ⁼ 576 + 49

     = 625

AC = 25

For angle A, the perpendicular is BC

(i) sin A = BC/AC = 7/25

CosA = AB/AC = 24/25

(ii) SinC = AB/AC = 24/25

     CosC = BC/AC = 7/25

Question 2:

tanP - cotR 

using the Pythagorean theorem

PR² = QR² +  PQ²

13x13 = 12x12 + QR²

169 - 144 = QR²

QR = 5 

tanP = QR/QP = 5/12

cotR = QR/PQ = 5/12

tanP - cotR = 5/12 - 5/12

=0

Question 3

Given Sin A = 3/4

Question4

Given that 15cotA = 8

Then Cot A = 8/15

Tan A = /CotA = 15/8

Tan A = perpendicular/Base

Hypotenuse² = perpendicular² + base²

                     =15² +8²

                     =225 +64 =279

Therefore, hypotenuse = 17

So, Sin A = perp/hyp = 15/17

SecA = hyp/base = 17/8

Question 5

Given that Sec θ = 13/12

Using Pythagoras' th.

Sec θ = 13/12

ð Hyp/base = 13/12

ð Perp = square root of (13²-12²)  = 5

ð Tanθ = perp/base = 5/12

ð Cot θ = 1/tanθ = 12/5

Sinθ = perp/hyp = 5/13

Cosecθ = 1/sinθ = 13/5

 

Question 6

Given that A and B are acute angles

CosA = CosB

Consider the two triangles APQ and BXY

AP/AQ = BX/BY   ---[1]

We can assume AP/AQ = BX/BY = k.

Then,

AP =kAQ

And, BX =kBY

Using Pythagoras' theorem for both triangles

Question 7

 Given: Cotθ = 7/8 

to find: (1-sinθ) (1+sinθ)/(1+cosθ) (1-cosθ) 

 It can be written as

 

(ii) cot^ θ = (7/8)^2 = 49/64

Question 8

Question 9

Question 10:

(i)  The value of TanA is not always less than 1  - False

(ii) SecA = 12/5 for some value of angle A = true

(iii) False

(iv) False; value of sine lies between -1 and 1

Exercise 8.2

(i) sin 60 cos 30 + cos 60 sin 30

Question 3

Question 4

(i) Sin(A + B) = SinA + SinB

Identity becomes true if it is true for each angle.

However, this is only true for zero degrees.

(ii) No, for the angle 0 to 90 degrees, it increases rather than decreases. 

(iii) No, Cosθ decreases with the increase in angle

(iv) No, Sinθ = Cosθ is only true for 45 degrees.

(v) True

-------------------Exercise 8.3 ...................................

Question 4(vii)