Table for Trigonometric Ratios for Specific Angle - Introduction to Trigonometry, CBSE, Class10, Mat | Extra Documents, Videos & Tests for Class 10 PDF Download

TRIGONOMETRICAL RATIO OF STANDARD ANGLES
 T-Ratios of 45°

Consider a ΔABC in which B = 90° and A = 45°.
Then, clearly, C = 45°.
 AB = BC = a (say).

T-Ratios of 60° and 30°
Draw an equilateral ΔABC with each side = 2a.

Then, A = B = C = 60°.
From A, draw AD  BC.

Then, BD = DC = a, BAD = 30° and ADB = 90°.

 T-Ratios of 30°

In ΔADB we have : ADB = 90° and BAD = 30°.

T-Ratios of 0° and 90°

T-Ratios of 0°
We shall see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC (see figure), till it becomes zero. As A gets smaller and smaller, the length of the side BC decreases. The point C gets closer to point B, and finally when A becomes very close to 0°, AC becomes almost the same as AB.

When A is very close to 0°, BC gets very close to 0 and so the value of  sin A =BC/ACis very close to O. Also,

when A is very close to 0°, AC is nearly the same as AB and so the value of cos A= AB/AC is very close to 1.

This helps us to see how we can define the values of sin A and cos A when A = 0°. We define :

sin 0° = 0 and cos 0° = 1.

Using these, we have :

T-Ratos of 90° till it'becomes 90°. As A gets larger and larger, C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when A is very close to 90°, C becomes very close to 0° and the side AC almost coincides with side BC (see figure).

When C is very close to 0°, A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when A is very close to 90°, C is very close to 0°, and the side AB is nearly zero,
so cos A is very close to 0. So, we define:

sin 90° = 1 and cos 90° = 0.

Using these, we have

Table for T-Ratios of Standard Angles

REMARK : (i) As θ increases from 0° to 90°, sin θ increases from 0 to 1.
(ii) As θ increases from 0° to 90°, cos θ decreases from 1 to 0.
(iii) As θ increases from 0° to 90°, tan θ increases from 0 to ∞.
(iv) The maximum value of 1/sec θ, 0° θ90⁰  is one.
(v) As cos θ decreases from 1° to 0, θ increases from 0 to 90°.
(vi) sin θ and cos θ can not be greater than one numerically.
(vii) sec θ and cosec θ can not be less than one numerically.
(viii) tan θ and cot θ can have any value.

COMPETITION WINDOW

T-RATIOS OF SOME ANGLES LESS THAN 90°

Ex.6 In ΔABC, right angled at B, BC = 5 cm, BAC = 30°, find the length of the sides AB and AC.
 Sol. We are given
BAC =30°, i.e., A = 30°

and BC = 5cm

  ...[∴ sin 30⁰ = 1/2] 

     ...[cos 30⁰ = √3/2]     

Ex.7 In ΔABC, right angled at C, if AC = 4 cm and AB = 8 cm. Find A and B.
 Sol. We are given, AC = 4 cm and AB = 8 cm

         ....[A + B = 90° ]

Now A = 90° – B                                                        
= 90° – 30° = 60°

Hence, A = 60° and B = 30°.

Ex.8 Find the value of θ in each of the following :

(i) 2 sin 2θ =

(ii) 2 cos 3θ = 1 

(iii)  tan 2θ – 3 = 0

Sol.

We have,

COMPETITION WINDOW

USING TRIGONOMETRIC TABLES

A Trigonometric Table consists of three parts :

(i)   A column on the extreme left containing degrees from 0° to 89°.
(ii) Ten columns headed by 0', 6', 12', 18', 24', 30', 36', 42', 48', and 54',
(iii) Five columns of mean differences, headed by 1', 2', 3', 4' and 5'. The mean differences is added in case of
sines, tangents and secants. The mean difference is subtracted in case of cosines, cotangents and cosecants.
The method of finding T-ratios of given angles using trigonometric tables, will be clear from the following
example :
Find the value of sin 43° 52'.
We have, 43°52' = 43°48' + 4'

In the table of natural sines, look at the number in the row aganist 43° and in the column headed 48' as shown
below.
From Table of Natural Sines :

TO FIND THE ANGLE WHEN ITS T-RATIO IS GIVEN

Find θ, when sin θ =0.7114.

From the table, find the angle whose sine is just smaller than 0.7114.
We have sinθ = 0.7114
Sin 45° 18' = 0.7108
Diff. = 0.0006

Mean difference of 6 corresponds to 3'.
 Required angle = (45° 18' + 3') = 45° 21'
Find θ, when cos θ = 0.5248
From the table, find the angle whose cosine is just greater than 0.5248
We have cosθ = 0.5248
cos 58° 18' = 0.5255
Diff. = 0.0007
And, 7 corresponds to 3'.
 Required angle = 58° 18' + 3' = 58° 21'

T-RATIOS OF COMPLEMENTARY ANGLES

Complementary Angles
Two angles are said to be complementary, if their sum is 90°.
Thus, θ° and (90° – θ) are complementary angles.
T-ratios of Complementary Angles
Consider ΔABC in which B = 90° and A = θ°.
C = (90° – θ).
Let AB = x. BC = y and AC = r.

Aid to memory
               Add co if that is not there
              Remove co if that is there
Thus we have,

In other words :
sin (angle) = cos (complement) ; 
cos (angle) = sin (complement)
tan (angle) = cot (complement) ;  
cot (angle) = tan (complement)
sec (angle) = cosec (complement) ;
cosec (angle) = sec (complement)

where complement = 90° – angle

Ex.9 Without using tables, evaluate:

Sol.

REMARK : (i) The above example suggests that out of the two t-ratios, we convert one in term of the t-ratio  of the complement.
                  (ii) For uniformity, we usually convert the angle greater than 45° in terms of its complement.

Ex.10 Without using tables, show that (cos 35° cos 55° – sin 35° sin 55°) = 0.

Sol. LHS = (cos 35° cos 55° – sin 35° sin 55°)
= [(cos 35° cos 55° – sin (90° – 55°) sin (90° – 35°)]
= (cos 35° cos 55° – cos 55° cos 35°) = 0 = RHS.
 [  sin (90° – θ) = cosθ and cos (90° – θ) = sin θ]