160. There is one point to be noticed in using the columns headed Diff. It has been pointed out that 2021 (at the top of the second column) means 0002021. Now the 790 (at the top of the eighth column) means not 000790, but '0000790. The rule is this; the right-hand figure of the Diff. must be placed in the seventh place of decimals and the requisite number of cyphers prefixed. Thus 9 means that the difference is '0000009, Diff. 161. Page 170 also gives the tabular logs. of ratios between 57° and 58°. Suppose we wanted L tan 57° 20′. We e now start with the line at the bottom of the page and run our eye up the column which has Tang. at its foot. We go up this column until we arrive at the number which is on the same level as the number 20 in the extreme right-hand column. This number we find to be 10∙1930286, which is therefore the value of and the angle whose L tan is 9·6195283. 5. Find the angle whose L cos is 9.993, given 6. Find the angle whose L sec is 10-15, given L sec 44° 55′ = 10.1498843, diff. for 1'=1260. 7. From the table on page 170 find the values of 8. With the help of the same page solve the equations (1) Ltan 0=10∙1959261, (3) L cos 0=9·9259283, (2) L cosec 0=10.0738125, and (4) L sin 0 = 9·9241352. 9. Take out of the tables L tan 16° 6′ 23′′ and calculate the value of the square root of the tangent. 10. Change into a form more convenient for logarithmic computation (i.e. express in the form of products of quantities) the quantities CHAPTER XII. RELATIONS BETWEEN THE SIDES AND THE TRIGONOMETRICAL RATIOS OF THE ANGLES OF ANY TRIANGLE. 162. IN any triangle ABC, the side BC, opposite to the angle A, is denoted by a; the sides. CA and AB, opposite to the angles B and C respectively, are denoted by b and c. 163. Theorem. In any triangle ABC, sin A sin B sin C a b C i.e. the sines of the angles are proportional to the opposite sides. Draw AD perpendicular to the opposite side meeting it, produced if necessary, in the point D. [If the angle C be obtuse, as in the second figure, we have In a similar manner by drawing a perpendicular from B upon CA we have sin C sin A If one of the angles, C, be a right angle as in the third figure we have sin C = 1, 164. In any triangle to find the cosine of an angle in terms of the sides. Let ABC be the triangle and let the perpendicular from A on BC meet it, produced if necessary, in the point D. First, let the angle C be acute, as in the left-hand figure. By Euc. II. 13, we have AB2 = BC2 + CA2 – 2BC . CD......... Secondly, let the angle C be obtuse, as in the right |