Difference between revisions of "1979 AHSME Problems/Problem 20"
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− | Since <math>a=\frac{1}{2}</math>, <math>b=\frac{1}{3}</math>. Now we evaluate <math>\arctan a</math> and <math>\arctan b</math>. Denote <math>x</math> and <math>\theta</math> such that <math>\arctan x = \theta</math>. Then <math>\tan(\arctan(x)) = \tan(\theta)</math>, and simplifying gives <math>x = \tan(\theta)</math>. So <math>a = \tan(\theta_a) = \frac{1}{2}</math> and <math>b = \tan(\theta_b) = \frac{1}{3}</math>. The question asks for <math>\theta_a + \theta_b</math>, so we try to find <math>\tan(\theta_a + \theta_b)</math> in terms of <math>\tan(\theta_a)</math> and <math>\tan(\theta_b)</math>. Using the angle addition formula for <math>\tan(\ | + | Since <math>a=\frac{1}{2}</math>, <math>b=\frac{1}{3}</math>. Now we evaluate <math>\arctan a</math> and <math>\arctan b</math>. Denote <math>x</math> and <math>\theta</math> such that <math>\arctan x = \theta</math>. Then <math>\tan(\arctan(x)) = \tan(\theta)</math>, and simplifying gives <math>x = \tan(\theta)</math>. So <math>a = \tan(\theta_a) = \frac{1}{2}</math> and <math>b = \tan(\theta_b) = \frac{1}{3}</math>. The question asks for <math>\theta_a + \theta_b</math>, so we try to find <math>\tan(\theta_a + \theta_b)</math> in terms of <math>\tan(\theta_a)</math> and <math>\tan(\theta_b)</math>. Using the angle addition formula for <math>\tan(\alpha+\beta)</math>, we get that <math>\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}</math>. Plugging <math>\tan(\theta_a) = \frac{1}{2}</math> and <math>\tan(\theta_b) = \frac{1}{3}</math> in, we have <math>\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in radians is <math>\boxed{\textbf{(C) } \frac{\pi}{4}}</math>. |
== See also == | == See also == |
Revision as of 11:51, 9 January 2017
Problem 20
If and then the radian measure of equals
Solution
Solution by e_power_pi_times_i
Since , . Now we evaluate and . Denote and such that . Then , and simplifying gives . So and . The question asks for , so we try to find in terms of and . Using the angle addition formula for , we get that . Plugging and in, we have . Simplifying, , so in radians is .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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