282. To solve a triangle when the three sides are given. [See Arts. 165, 166.] We have b2 + c2 - a2 COS A = 2bc The factor b2 + c2 – a2 has here to be calculated by multiplications: hence this formula is not well adapted for logarithmic calculations. On the other hand, the formula log tan 4 = log (s - b) + log (sc) - log (sa) - logs. -a, s-b, 8-c If all the angles are to be found the tan formula is the most convenient, because only the four logarithms of s, s will be required for the three angles. Example. Given a=35742 yards, b=29813 yards, c=47265 yards, find A. 3403191logs + log (sa)-log (8-b)-log (sc). .. L tan 410–3403191=9'6596809. Now 9.6597076 L tan 24° 33′ 9.6593733 L tan 24° 32′ 0003343 diff. for l' .. 0003076=diff. for 19 of 60′′, i.e. 54′′ ·5, .. 4 = 24° 32′ 54′′ ·5, ... A=49° 5′ 49′′. 283. To solve a triangle when two sides and the included angle are given. [See Arts. 167, 168.] I. To find a, we have a2 = b2 + c2 - 2bc cos A. This is not well adapted to logarithmic calculation. may, however, transform it as follows: a2 = b2+c2 2bc (1-2 sin2 A), Now there must be some angle, say, whose tangent Ө We By the help of the tables this angle may be found. Thus since Having found B and C we may find a from either of the equations or a = (b−c) cos 1⁄2 A cosec 1⁄2 (B – C). The two latter of these (which the student may easily prove) have the advantage over the first that they require only two new logarithms. Thus log a = log (b+c) + log sin A + log sec † (B − C'). Example. Given b=723, c= =259, A=35° 18', solve the triangle. Then I tan 6=log 2+ log blog c-log (b-c) + L sin 17° 39′ =·3010300+1·4295691+1.2066499 ...a=533·0614=533 06 approximately. Next L tan(B-C)=log (b-c) - log (b+c) + L cot A (B-C)=56° 2′ 39", and (B+C)= 72° 21'′, =2.6665180-2.9921115+10:4973279 =10.1717344, ... B=128° 23′ 39′′ and C=16° 18′ 21′′. Hence log a = log (b+c)+ log sin 17° 39′+ log sec 56° 2′ 39′′, =2.9921115+1·4817315+2529351, = 2.7267781, .. a=533·0625=533'06 approximately (as above). 284. To solve a triangle when two sides and a non-included angle are given. [See Arts. 169, 170.] .. L sin C = log c – log b + L sin B. This formula is immediately adapted to logarithmic calculation and need not be further discussed. EXAMPLES XI. 1. Prove from definition (the indices being positive in 2. Interpret ao, a-5, a, at; and write down the logarithms. of each expression to the base ɑ. 3. Find the values of 3-4, 27, 128-4, 219o. 4. Express the identities in Ex. 1, in terms of powers instead of roots of a. and write down the logarithms of each expression to the base x. 6. 7. Find the values of log, 243, log, 343, log1 64, log10 10000, log, 81, logi 5/11, log √27, log, /, log 125, 1 log1s, log, 2, log 3, log, 81, log, 3/2, log,,log10 1 3/5 logs logo 00001, log, cos 60°, log, cos 45°, Find the characteristics of log10 3245, log10 123, log10 3·45, log10 ·1234, Prove the following statements: (8-23). 16. = If loga x = log y = log. z, then each = log xPyr to the base arbic. 17. log x to base a2 = log /x to base a. 18. 2 log cos A = log (1 + sin A) + log (1 − sin 4). 19. 2 log tan A = log (sec A-1)+ log (sec A + 1). 20. log sin 24 = log 2 + log cos A + log sin A. 21. log cos 24 = log (cos A + sin A) + log (cos A – sin 4). 22. log (1 + sin 24) = 2 log (cos A + sin A). 23. log tan (4+ 45°) = log (1 + tan 4) - log (1 - tan A). Logarithms to base 10. 24. Given log 2 = 3010300, find the logarithms of 25. 1 4, 8, 128, 2/2, 1024, §, 164, 25, 03125, 200, -002. Given log 23010300, find the logarithms of 26. Given log 34771213, find the logarithms of 9, 1, 30, 243, 03, ∙1. 27. Given log 2 and log 3 (as above) find the logs of 6, 18, 15, 135, 144, 750, 2, 004. 28. 29. Given log 7 = 8450980, find the logs of 343, 343000, 343, 00343, 3.43. Given log 2, log 3, and log 7 (as above) find the logs of 35, 210, 245, 28, 42, 63, 441, 504. State in each case the possible error in the result calculated. |