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
***********
No comments:
Post a Comment