Difference between revisions of "1979 AHSME Problems/Problem 20"
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− | == Problem | + | == Problem == |
If <math>a=\tfrac{1}{2}</math> and <math>(a+1)(b+1)=2</math> then the radian measure of <math>\arctan a + \arctan b</math> equals | If <math>a=\tfrac{1}{2}</math> and <math>(a+1)(b+1)=2</math> then the radian measure of <math>\arctan a + \arctan b</math> equals | ||
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\textbf{(E) }\frac{\pi}{6}</math> | \textbf{(E) }\frac{\pi}{6}</math> | ||
− | == | + | ==Solution== |
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
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>. | 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>. | ||
+ | |||
+ | == Solution 1.1 (Guiding through the thought process) == | ||
+ | Thinking through the problem, we can see <math>\arctan a</math> is the angle whose tangent is <math>0.5</math>. <math>\arctan b</math> is the angle whose tangent is <math>\frac{1}{3}</math>. Call the former angle <math>\theta_1</math>, the latter <math>\theta_2</math>. So we are trying to find <math>\theta_1+\theta_2</math>. So given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calculators). Plug in as above and continue on. | ||
+ | |||
+ | ~hastapasta | ||
== See also == | == See also == |
Latest revision as of 13:33, 16 July 2024
Problem
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
.
Solution 1.1 (Guiding through the thought process)
Thinking through the problem, we can see is the angle whose tangent is
.
is the angle whose tangent is
. Call the former angle
, the latter
. So we are trying to find
. So given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calculators). Plug in as above and continue on.
~hastapasta
See also
1979 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 |
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