LOGARITHMIC COSINES

Method of using the table of sines : this table is also known as the table of logarithmic cosines . At intervals of 60'(or 1°), we can find the values of cosines angles ranging from 0° to 89°.

         The table of logarithmic cosines is essentially divided into the following (parts) sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

       (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

                        OR

 at intervals of 0.1° , the horizontal row at the top of the chart is from 0.0° to 0.9° .

       (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column ( Subtract Mean Difference).

* Note: 

           (i) we obtain the cosines value of a given angle from the chart, which is expressed in five decimal places. 

           (ii)  The table is also drawn in such a way that we can determine the logarithmic cosine values of any given angle between 0° and 89°.

             (iii) sine this portion of the chart is referred to as the subtract mean difference column.

            (iv) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5' in contain not available 86° to 89° column.

SOLVED EXAMPLES USING THE TABLE OF LOGARITHMIC COSINES :

1. Find the value of LOG ( cos 60°) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 60°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.6990

   LOG ( cos 60°) ≈ "bar" 1.6990 ... Ans

2. Find the value of LOG ( cos19°30' ) ≈ ? 

   ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 19°.

 Then we move horizontally to the right at the top of the column headed by 30' (or 0.5°) and read the figure "bar" 1.9743

   LOG ( cos19°30' ) ≈  "bar" 1.9743 ... Ans

3. Find the value of LOG ( cos 58°9’) ≈ ? 

     ⇒SOLUTION : 

LOG ( cos 58°9’) ≈LOG ( cos 58°6' + 3’) ≈

Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 58°.

 Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.7230

Now we move further right along the horizontal line of angle 58° to the column headed by 3' of mean difference and read the figure 7 there; this figure of the table does not contain decimal sign. In fact, 7 implies 0.0007 .                      

LOG ( cos 58°6' + 3’) ≈ (bar 1.7230 - 0.0007)

                                     ≈ "bar" 1.7223

 LOG ( cos 58°9’) ≈  "bar" 1.7223 ... Ans

4. Find the value of LOG ( cos 67°32’) ≈ ? 

   ⇒SOLUTION : 

LOG ( cos 67°32’) ≈LOG ( cos 67°30' + 2’) 

      Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 67°.

      Then we move horizontally to the right at the top of the column headed by 30' (or 0.5°) and read the figure "bar" 1.5828

       Now we move further right along the horizontal line of angle 67° to the column headed by 2' of mean difference and read the figure 6 there; this figure of the table does not contain decimal sign. In fact, 6 implies 0.0006 .  

  LOG ( cos 67°30' + 2’) ≈ (bar 1.5828 - 0.0006) 

                                        ≈ "bar" 1.5822

LOG ( cos 67°32’) ≈ "bar" 1.5822 ... Ans                  

5. Find the value of LOG ( cos 72°8’) ≈ ? 

     ⇒SOLUTION :

   LOG ( cos 72°8’) ≈ LOG (cos 72°6’+ 2') ≈

         Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 72°.

       Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.4876 

       Now we move further right along the horizontal line of angle 72° to the column headed by 2' of mean difference and read the figure 8 there; this figure of the table does not contain decimal sign. In fact, 8 implies 0.0008  . 

 LOG (cos 72°6’+ 2') ≈"bar"(1.4876 - 0.0008)

                                   ≈ "bar"1.4868

  LOG ( cos 72°8’) ≈ "bar"1.4868  .... Ans

                            **********

LOGARITHMIC SINES

      Method of using the table of sines  : this table is also known as the table of logarithmic sines . At intervals of 60'(or 1°), we can find the values of sines angles ranging from 0° to 89°.

         The table of logarithmic sines is essentially divided into the following (parts) sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

       (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

       (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column ( Add Mean Difference).

* Note: 

           (i) we obtain the sine value of a given angle from the chart, which is expressed in five decimal places. 

           (ii) we know that the sine of any given angle is equal to that of the cosine of its complementary angle, i.e., sin(θ) =cos(90º - θ) . The table is also drawn in such a way that we can determine the logarithmic sin values of any given angle between 0° and 89°.

             (iii) sine this portion of the chart is referred to as the add mean difference column.

            (iv) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5' in contain not available 0° to 3° column.

SOLVED EXAMPLES USING THE TABLE OF LOGARITHMIC SINES :

1. Find the value of  LOG ( sin 60°) ≈  ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 60°. 

           Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.9375 ,Therefore LOG ( sin 60°) ≈ "bar" 1.9375   ... Ans

2. Find the value of LOG ( sin 19.4°) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 19°.

 Then we move horizontally to the right at the top of the column headed by 24' (or 0.4°) and read the figure "bar" 1.5213

       LOG ( sin 19.4°) or ≈ LOG ( sin 19°24') 

                                 ≈ "bar" 1.5213 ... Ans

3. Find the value of  LOG ( sin 32°5’) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 32°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.7242 , 

Now we move further right along the horizontal line of angle 32° to the column headed by 5' of mean difference and read the figure 10 there; this figure of the table does not contain decimal sign. In fact, 10 implies 0.0010. 

        LOG ( sin 32°5’) ≈ LOG ( sin 32°0’+5')

                          ≈  ( bar 1.7242 + 0.0010 ) 

                           ≈  "bar" 1.7252 ... Ans

4. Find the value of  LOG ( sin 72°60’) ≈  ? 

    ⇒SOLUTION :  LOG ( sin 72°60’) 

          or ≈ LOG ( sin 73°)      ( •.• 60' = 1°) 

 Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 73°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.9806 

.•.  LOG ( sin 72°60’)  ≈ "bar" 1.9806 .. Ans

5. Find the value of  LOG ( sin 67°8’) ≈ ? 

    ⇒SOLUTION : 

    LOG ( sin 67°8’) ≈  LOG ( sin 67°6’ + 2')

   Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 67°.

 Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.9643 , 

Now we move further right along the horizontal line of angle 67° to the column headed by 2' of mean difference and read the figure 1 there; this figure of the table does not contain decimal sign. In fact, 1 implies 0.0001 .                      

               ≈ ( bar 1.9643 + 0.0001) 

               ≈ "bar" 1.9644

    

                         ***********

RECIPROCALS OF FOUR-FIGURE NUMBERS

Introduction : 

          Reciprocal is simply defined as the inverse of a number Or a value i.e. the reciprocal of a number is 1 divided by that number. If n is a real number, then its inverse will be 1/n or n⁻¹. A reciprocal is also a number taken to the power of -1.

For Example , the reciprocal (or inverse) of 5 is 1 divided by 5 , which is 1/5 or 5⁻¹.

1/5 or 5⁻¹ = 0.2 ( from chart 0.2000 four- figure number )

         The table of reciprocal numbers is also known as the table of reciprocal four-figure numbers. We can find reciprocal with values ranging from 1.0 to 9.9 at intervals of 0.1. The table of reciprocal four- figure numbers is essentially divided into three sections, as shown in the following tables. 

(i) at intervals of 0.1, the number in the table's extreme left vertical column ranges from 1.1 to 9.9. 

(ii) the horizontal row at the top of the chart is from 0 to 9 at intervals 1. 

(iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean differences column.

Find the reciprocal of following numbers, using the reciprocal four-figure numbers chart.  

  If x  ≥ 1

(i) 1/1.722 = ? 

⇒ SOLUTION :  The number in the reciprocal four-figure numbers table extreme left vertical column in 1.7 ,the horizontal row at the top of the chart in 2 consist of containing four digits number 0.5814 and horizontal row at the top of the chart in number 2 consist of containing one digit number 7 [= 0.0007 mean difference subtract ]

    1/1.722 = 0.5814 - 0.0007 

                   = 0.5821   .... Ans

  

(ii) 1/17.22 =  ? 

⇒SOLUTION : 1/1.722 x 10¹(decimal place change  right to left and multiply  10¹ )

      = ( 1/1.722 ) x 10⁻¹  (·.· 1/10¹ = 10⁻¹ )

      = 0.5821 x 10⁻¹ 

      = 0.05821     .....  Ans

 

(iii) 1/172.2 =  ? 

⇒SOLUTION : 1/1.722 x 10² (decimal place change from right to left and multiply  10² )

     = ( 1/1.722 ) x 10⁻²   ( ·.· 1/10² = 10⁻² )

     = 0.5821 x 10⁻² 

     = 0.005821    ..... Ans

 

 

(iv) 1/1722 =  ? 

⇒SOLUTION : 1/1.722 x 10³ (decimal place change from  right to left and multiply x 10³ )

    = ( 1/1.722 ) x 10⁻³ ( ·.· 1/10³ = 10⁻³ )

    = 0.5821 x 10⁻³ 

    = 0.0005821  ..... Ans

  

Find the reciprocal of following numbers, using the reciprocal four-figure numbers chart.  

 If x < 1

(i) 1/0.1722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻¹ (decimal place change from left to right and multiply x 10⁻¹ )

      = ( 1/1.722 ) x 10¹    ( ·.· 1/10⁻¹ = 10¹ )

      = 0.5821 x 10¹ 

      = 5.821   ...... Ans

(ii) 1/0.01722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻² (decimal place change from  left to right  and multiply x 10⁻² )

     = ( 1/1.722 ) x 10²   ( ·.· 1/10⁻² = 10² )

     = 0.5821 x 10² 

     = 58.21      ..... Ans

(iii) 1/0.001722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻³(decimal place change from left to right and multiply x 10⁻³ )

     = ( 1/1.722 ) x 10³  ( ·.· 1/10⁻³ = 10³ )

     = 0.5821 x 10³

     = 582.1     .....  Ans

(iii) 1/0.001722 =  ? 

⇒ SOLUTION : 1/1.722 x 10⁻⁴ (decimal place changes from left to right and multiply x 10⁻⁴  )

    = ( 1/1.722 ) x 10⁴    ( ·.· 1/10⁻⁴ = 10⁴ )

    = 0.5821 x 10⁴ 

    = 5821     ......  Ans

  

                      *********

SQUARE ROOTS(FROM 1 to 10 AND 10 TO 99)

By MR. DEVIDAS MADHAO MISTARI

                            M.Sc. (physics) 

                          Material science

 Introduction :      

           A number is a square root that, when multiplied by itself, corresponds to the desired value. So, for example, the square root of x = 64 is 8 ( 8x8 = 64 ). Squaring is the act of multiplying a number by itself.

        [ Using square root symbol :  √(x)  ] where, x is number. 

  

           Method of using the table of square roots (from 1 to10 ) : this table is also known as the table of square roots . At intervals of 0.1, we can find the values of square roots  ranging from 1.0 to 9.9.

         The table of square roots is essentially divided into the following parts (three parts) , as shown in the following tables. 

        

       (i) the number in the table's extreme left vertical column are from 1.1 to 9.9 at intervals of 0.1 . 

 

        (ii) at intervals of 1 , the horizontal row at the top of the chart is from 0 to 9.

    

       (iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean difference column.

      * NOTE :  The power of 10 extracted under the root sign must always be even number. 10⁻²,10⁻⁴,10² and 10⁴ ⇒ Even power (-2),(-4),(2) and (4)

1. Square roots (from 1 to 10) 

      For Example :

   

     ( i )   √(6.479) = ? 

SOLUTION : The number in the square roots (from 1 to 10) table extreme left vertical column in 6.4 ,the horizontal row at the top of the chart in 7 consist of containing four digits number 2.544 and horizontal row at the top of the chart in number 9 consist of containing one digit number 2 = 0.002 (mean difference add)·

  √(6.479) = ( 2.544 + 0.002 ) = 2.546 (·.·1 < 5 < 9 )

Therefore √(6.479) = 2.546  ... Answer

     

     ( ii )  √(647.9) =  ? 

     SOLUTION : √(6.479 x 10²)   (decimal place changes from 3ʳᵈ number to 2ⁿᵈ number digit right to left and multiply x 10² )

 = √(6.479)  x 10¹   [ ·.· √(10²) = 10¹ ]

 = ( 2.544 + 0.002 ) x 10¹

 = 2.546 x 10¹ = 25.46

Therefore √(6.479) = 25.46 ... Answer

   ( iii )  √(0.06479) =  ? 

   SOLUTION : √(6.479 x 10⁻² )   (decimal place changes from 1ˢᵗ to 2ⁿᵈ left to right and multiply x 10⁻² )

   = ( 2.544 + 0.002 ) x 10⁻¹

            [·.· √(10⁻²)=10⁻¹] 

   = 2.546 x 10⁻¹ = 0.2546

   ( iv )  √(0.0006479) = ? 

SOLUTION : √(6.479 x 10⁻⁴)   (decimal place changes from 1ˢᵗ to 4ᵗʰ left to right and multiply x10⁻⁴)

 = ( 2.544 + 0.002 ) x 10⁻²  

         [·.· √(10⁻⁴)=10⁻²]

 = 2.546 x 10⁻² = 0.02546

        Method of using the table of square  (from 10 to 99 ): this table is also known as the table of square roots . At intervals of 1, we can find the values of square roots  ranging from 10 to 99.

         The table of square roots is essentially divided into the following parts, as shown in the following tables. 

        

       (i) the number in the table's extreme left vertical column are from 10 to 99 at intervals of 1 . 

 

        (ii) at intervals of 1 , the horizontal row at the top of the chart is from 0 to 9.

    

       (iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean difference column.

2. Square roots (from 1 to 99) 

      For Example :

 ( i ) √(64.79) = ? 

SOLUTION  : √(64.79) = (8.044 + 0.006) 

  = 8.050   The number in the square roots (from 10 to 99) table extreme left vertical column in 64 ,the horizontal row at the top of the chart in 7 consist of containing four digits number 8.044 and horizontal row at the top of the chart in number 9 consist of containing one digit number 6 = 0.006 (mean difference add)·

   Therefore √(64.79) = 8.050

  

( ii )  √(6479) = ? 

  SOLUTION :  √(64.79 x 10²) (decimal place changes from 4ᵗʰ number to 2ⁿᵈ number digit right to left and multiply x 10² )

    = (8.044 + 0.006) x 10¹

            [·.· √(10²)=10¹] 

    = 8.050 x 10¹ = 80.50

   

 ( iii )  √(0.6479) = ? 

SOLUTION : √(64·79 x 10⁻²) (decimal place changes from 1ˢᵗ  to 2ⁿᵈ left to right and multiply x 10⁻² )

= (8.044 + 0.006) x 10⁻¹     

     [·.· √(10⁻²)=10⁻¹] 

= 8.050x 10⁻¹ = 0.8050

   

  ( iv )  √(0.006479) = ? 

SOLUTION  : √(64·79 x 10⁻⁴) (decimal place changes from 1ˢᵗ to 4ᵗʰ right to left and multiply x 10⁻⁴ )

= (8.044 + 0.006) x 10⁻²   

    [·.· √(10⁻⁴)=10⁻²] 

= 8.050 x 10⁻² = 0.08050

   

SQUARES

By MR. DEVIDAS MADHAO MISTARI

                                             M.Sc. (physics) 

                                            Material science

HOW TO USE SQUARES CALCULATIONS WITH ILLUSTRATIONS . 

          

The Squares table consists of two parts:

1. The extreme left column consisting of one to two digit numbers from 0 to 99.

2. Ten right columns containing one to six digit numbers , headed by the digits from 0 to 9.

     * Note : From the identity, exact squares of four figure numbers can be easily calculated.

               ( x + y)² = x² + 2xy +y²

              ( x - y)² = x² - 2xy + y²

A Square table is shown below:

    

Solved examples using the table of  squares : 

 1. Find the square of one digit number   (i) 3²   (ii) 5²   (iii) 9²    (iv) 8²

  Solution :  Easily find the square of one digit number and remember in mind.One digit numbers 1 to 9.

         (i) 3² = 9  →Solution :  Find the value of 3 by the using the squares table . Then we move horizontally to the right at the top of the column headed by 3 and the extreme left column consisting of  number 0  read the figure 9. Therefore 3 squares is 9.

        (ii) 5² = 25  →Solution : Find the value of 5 by the using the squares table . Then we move horizontally to the right at the top of the column headed by 5 and the extreme left column consisting of number 0 read the figure 25. Therefore 5 squares is 25.

       (ii) 8² = 64 → Solution : Find the value of 8by the using the squares table . Then we move horizontally to the right at the top of the column headed by 8 and the extreme left column consisting of number 0 read the figure 64. Therefore 8 squares is 64.

      (iv) 9² = 81 →Solution : The extreme left column consisting of number 0. Right columns containing read two digit number 81, headed by the digits 9 .Therefore 9 squares is 81.

2. Find the square of two digit number Quickly find the square of two digit number using squares table . Two digit numbers 10 to 99. 

            For example  ⇒

           (i) 10² = 100 →Solution :Determine the value of squres easily find the square of 10 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 0 and Right columns containing read three digit number 100.

          (ii) 12² = 144 →Solution : Determine the value of squres easily find the square of 12 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 2 and Right columns containing read three digit number 144.

         (iii) 15² = 225 →Solution : Find the value of squres easily find the square of 15 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 5 and Right columns containing read three digit number 225.

         (iv) 98² = 9604→Solution : Determine the value of squres quickly find the square of 98 using the squares table. The extreme left column consisting of number 9. Then we move horizontally to the right at the top of the column headed by 8 and Right columns containing read four significant figure number 9604.

3. Find the square of three digit number quickly find the square of three digits numbers using squares table . Three digits numbers 100 to 999.  

           For example  ⇒

          (i) 103² = 10609 →Solution : The extreme left column consisting of number 10. Then we move horizontally to the right at the top of the column headed by 3 and Right columns containing read five digit number 10609.

          (ii) 121² = 14641 →Solution : The extreme left column consistingSolution of number 12. Then we move horizontally to the right at the top of the column headed by 1 and Right columns containing read five digit number 14641.

         (iii) 125² = 15625 → Solution : The extreme left column consisting of number 12. Then we move horizontally to the right at the top of the column headed by 5 and Right columns containing read five digit number 15625.

         (iv) 888² = 788544 → Solution : The extreme left side column consisting of number 88. Then we move horizontally to the right at the top of the column headed by 8 and Right columns containing read six significant figures 788544.

         (v)  987² = 974169 → Solution : The extreme left side column consisting of number 98. Then we move horizontally to the right at the top of the column headed by 7 and Right columns containing read six significant digits 974169.

  ****************************************

NATURAL TANGENTS

HOW TO USE NATURAL TANGENTS CALCULATIONS WITH ILLUSTRATIONS

Method of using the table of  tangents : this table is also known as the table of natural tangents .At intervals of 60', we can find the values of sines and cosines of angles ranging from 0° to 89°.

         The table of natural tangents is essentially divided into the following sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

        (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

        (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column.

* Note: 

            we obtain the tangent value of a given angle from the chart, which is expressed in four decimal places. 

Solved examples using the table of natural tangents : 

         1. Find the value of tan 55°. 

Solution:   Find the value of sin 55° by the using the table of natural sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 55°.               

               Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure 1.4281, which is the require value of tan 55°. 

Therefore tan55° = 1.4281   .... Answer

      2. Find the value of tan 60°36’

Solution:  Find the value of tan 60°36’ by the using the table of natural tan we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 60°. 

           Then we move horizontally to the right at the top of the column headed by 36' (or 0.6°) and read the figure 1.7747 , which is the require value of tan 60°36’ .

 Therefore tan 60°36’  = 1.7747 ... Answer

         3.Use tables to find tan72°60' 

Solution: Using the trigonometric table of natural tangents. tan 72°60' = tan ( 72° + 60') = tan (72°+ 1° ) = tan (71°)  Because 60' = 1°

Therefore  tan 70°60' = tan (71°)= 2.9042

 tan 72°60'= 2.9042     ... Answer

          4. Find the angle θ, tan θ = 1.4994

solution : tan θ = 1.4994

Therefore   θ = tan⁻¹ 1.4994 = 56°18′ because in the table, the value 1.4994 corresponds to the column of 18′ in the row of 56°.

θ = tan⁻¹ 1.4994 

    = 56°18′ or 56.3° .... Answer

NATURAL SINES AND NATURAL COSINES

HOW TO USE NATURAL SINES AND NATURAL COSINES CALCULATIONS WITH ILLUSTRATIONS

 

   Most important trigonometry formulae :

          sin(90º - θ) = cos(θ)

          sin(90º + θ) = cos(θ)

          cos(90º - θ) = sin(θ)

          cos(90º + θ) = -sin(θ)

          sin(180º - θ) = sin(θ)

          sin(180º + θ) = -sin(θ)

          cos(180º - θ) = -cos(θ)

          cos(180º + θ) = -cos(θ)

           where,θ is angle

      1º = 60´ i.e one degree equal sixty minute. 

 ( ºsymbol of degree  and  ´symbol of minute )

           0.1º =  6´  → ( 0.1º x 60´ =  6´ )

           0.2º = 12´ → ( 0.2º x 60´ =  12´ )

           0.3º = 18´ → ( 0.3º x 60´ =  18´ )

           0.4º =  24´→ ( 0.4º x 60´ =  24´ )

           0.5º = 30´ → ( 0.5º x 60´ =  30´ )

           0.6º = 36´ → ( 0.6º x 60´ =  36´ )

           0.7º = 42´ →  ( 0.7º x 60´ =  42´ )

           0.8º = 48´ →  ( 0.8º x 60´ =  48´ )

           0.9º = 54´ →  ( 0.9º x 60´ =  54´ )

           

      Method of using the table of sines and cosines: this table is also known as the table of natural sines and natural cosines.At intervals of 60', we can find the values of sines and cosines of angles ranging from 0° to 89°.

         The table of natural sines and natural cosines is essentially divided into the following sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

        (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

        (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column.

* Note: 

           (i) we obtain the sine or cosine value of a given angle from the chart, which is expressed in four decimal places. 

           (ii) we know that the sine of any given angle is equal to that of the cosine of its complementary angle, i.e., sin(θ) =cos(90º - θ) . The table is also drawn in such a way that we can determine the sin and cosine values of any given angle between 0° and 89°.

             (iii) Cosine this portion of the chart is referred to as the subtract mean difference column.

Solved examples using the table of natural sines : 

         1. Find the value of sin 44°. 

Solution:   Find the value of sin 44° by the using the table of natural sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 44°.               

               Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure 0.6947, which is the require value of sin 44°. Therefore sin 44° = 0.6947

          2. Find the value of sin 62°24’

Solution:  Find the value of sin 62°24’ by the using the table of natural sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 62°. 

           Then we move horizontally to the right at the top of the column headed by 24' (or 0.4°) and read the figure 0.8862, which is the require value of sin 62°24’ .

        Therefore sin 62°24’  = 0.8862

            3. Using the trigonometric table, find the value of sin 62°28'

Solution: To find the value of sin 62°28' by the using the trigonometric table of natural sines we need to first find the value of sin 62°24'.

           To find the value of sin 62°24’ by the using the table of natural sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 62°. 

           Then we move horizontally to the right at the top of the column headed by 24' (Or 0.4°)and read the figure 0.8862, which is the require value of sin 62°24'. 

       Therefore, sin 62°24' = 0.8862

         Now we move further right along the horizontal line of angle 62° to the column headed by 4' of mean difference and read the figure 5 there; this figure of the table does not contain decimal sign. In fact, 5 implies 0.0005. Now we know that when the value of an angle increases from 0° to 89°, its sine value increases continually from 0 to 1. Therefore, to find the value of sin 62°24' we need to add the value corresponding to 4’ with the value of sin 62°24'.                             

   Therefore, sin 62°28' = sin ( 62°24' + 4') = 0.8862 + 0.0005 = 0.8867 

             4.Use tables to find sin77°78' 

Solution: Using the trigonometric table of natural sine  sin77°78' = sin(77°60' + 28')                             But 60' = 1°

Therefore  sin77°78' = sin(78° + 28' ) = sin(78°+ 28') = sin(78°24' + 4')= 0.9796 + 0.0002 = 0.9798  

 sin77°78' = 0.9798     ... Answer

          5. Find the angle θ, sin θ = 0.9298

solution : sin θ = 0.9298 

Therefore   θ = sin⁻¹ 0.9298 = 68° 24′ because in the table, the value 0.9298 corresponds to the column of 24′ in the row of 68°.

   θ = sin⁻¹ 0.9298 = 68° 24′ .... Answer

Solved examples using the table of natural cosines : 

   1. Find the value of cos 63°28' ,Using the trigonometric table. 

solution:  The value of cos 63°28' = cos ( 63°24' + 4' ) = ( 0.4478 - 0.0010 mean difference subtract ) = 0.4468 Therefore  cos 63°28' = 0.4468  ... Ans

   2. Find the value of cos 25°.  

 Solution: Find the value of cos 25° by the using the table of natural cosines.  The extreme left column consisting of 0° to 89° and move downwards till we reach the angle 25°.    

         We then move horizontally to the right at the top of the column headed by 0' (or 0.0°) and say the figure 0.9063, which is the demand value of 25°.      

Therefore cos 25° = 0.9063 ... Answer

    3. Find the value of cos 28.9°.

  Solution: The value of cos 28.9°, using the natural cosine table cos 28.9° = cos (28° + 0.9° ) = cos 28° 54' = 0.8755 Therefore , cos 28.9°= 0.8755.... Ans