2007 iTest Problems/Problem 17

Problem

If $x$ and $y$ are acute angles such that $x+y=\frac{\pi}{4}$ and $\tan{y}=\frac{1}{6}$ , find the value of $\tan{x}$ .

$\text{(A) }\frac{37\sqrt{2}-18}{71}\qquad \text{(B) }\frac{35\sqrt{2}-6}{71}\qquad \text{(C) }\frac{35\sqrt{3}+12}{33}\qquad \text{(D) }\frac{37\sqrt{3}+24}{33}\qquad$

$\text{(E) }1\qquad \text{(F) }\frac{5}{7}\qquad \text{(G) }\frac{3}{7}\qquad \text{(H) }6\qquad$

$\text{(I) }\frac{1}{6}\qquad \text{(J) }\frac{1}{2}\qquad \text{(K) }\frac{6}{7}\qquad \text{(L) }\frac{4}{7}\qquad \text{(M) }\sqrt{3}\qquad$

$\text{(N) }\frac{\sqrt{3}}{3}\qquad \text{(O) }\frac{5}{6}\qquad \text{(P) }\frac{2}{3}\qquad \text{(Q) }\frac{1}{2007}\qquad$

Solution

From the second equation, we get that $y=\arctan\frac{1}{6}$ . Plugging this into the first equation, we get:

$x+\arctan\frac{1}{6}=\frac{\pi}{4}$

Taking the tangent of both sides,

$\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1$

From the tangent addition formula, we then get:

$\frac{{\tan{x}+\frac{1}{6}}}{{1-\frac{1}{6}\tan{x}}}=1$

$\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}$ .

Rearranging and solving, we get

$\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}$

Solution 2

We will use complex numbers: $y$ represent the complex number $6+i$ , and $x+y=\frac{\pi}{4}$ represents the fact $x\cdot y=1+i$ , so dividing $1+i$ by $6+i$ , we get $\frac{7+5i}{56}$ , which means $\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}$

2007 iTest (Problems, Answer Key)
Preceded by: Problem 16	Followed by: Problem 18
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

Page

Toolbox

Search

2007 iTest Problems/Problem 17

Contents

Problem

Solution

Solution 2

See Also