Difference between revisions of "1987 AHSME Problems/Problem 20"
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\textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad | \textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad | ||
\textbf{(D)}\ 1\qquad | \textbf{(D)}\ 1\qquad | ||
| − | \textbf{(E)}\ \text{none of these} </math> | + | \textbf{(E)}\ \text{none of these} </math> |
| + | |||
| + | ==Solution== | ||
| + | Because <math>\tan x \tan (90^\circ - x) = \tan x \cot x = 1</math>, <math>\tan 45^\circ = 1</math>, and <math>\log a + \log b = \log {ab}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math> | ||
== See also == | == See also == | ||
Revision as of 00:07, 17 April 2015
Problem
Evaluate
Solution
Because 
, 
, and 
, the answer is 
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
