Difference between revisions of "1979 AHSME Problems/Problem 20"

(Problem 20)
(SOLUTION)
 
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\textbf{(E) }\frac{\pi}{6}</math>
 
\textbf{(E) }\frac{\pi}{6}</math>
  
==SOLUTION==
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==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  

Latest revision as of 15:51, 17 June 2021

Problem #20

If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals

$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$

Solution

Solution by e_power_pi_times_i

Since $a=\frac{1}{2}$, $b=\frac{1}{3}$. Now we evaluate $\arctan a$ and $\arctan b$. Denote $x$ and $\theta$ such that $\arctan x = \theta$. Then $\tan(\arctan(x)) = \tan(\theta)$, and simplifying gives $x = \tan(\theta)$. So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$. The question asks for $\theta_a + \theta_b$, so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$. Using the angle addition formula for $\tan(\alpha+\beta)$, we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$. Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$. Simplifying, $\tan(\theta_a + \theta_b) = 1$, so $\theta_a + \theta_b$ in radians is $\boxed{\textbf{(C) } \frac{\pi}{4}}$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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