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
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