Difference between revisions of "2007 iTest Problems/Problem 17"
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If <math>x</math> and <math>y</math> are acute angles such that <math>x+y=\frac{\pi}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>. | If <math>x</math> and <math>y</math> are acute angles such that <math>x+y=\frac{\pi}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>. | ||
| + | |||
| + | <math>\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</math> | ||
| + | |||
| + | <math>\text{(E) }1\qquad | ||
| + | \text{(F) }\frac{5}{7}\qquad | ||
| + | \text{(G) }\frac{3}{7}\qquad | ||
| + | \text{(H) }6\qquad</math> | ||
| + | |||
| + | <math>\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</math> | ||
| + | |||
| + | <math>\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</math> | ||
== Solution == | == Solution == | ||
| − | + | From the second equation, we get that <math>y=\arctan\frac{1}{6}</math>. Plugging this into the first equation, we get: | |
| + | |||
| + | <math>x+\arctan\frac{1}{6}=\frac{\pi}{4}</math> | ||
| + | |||
| + | Taking the tangent of both sides, | ||
| + | |||
| + | <math>\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1</math> | ||
| + | |||
| + | From the tangent addition formula, we then get: | ||
| + | |||
| + | <math>\frac{{\tan{x}+\frac{1}{6}}}{{1-\frac{1}{6}\tan{x}}}=1</math> | ||
| + | |||
| + | <math>\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}</math>. | ||
| + | |||
| + | Rearranging and solving, we get | ||
| + | |||
| + | <math>\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | We will use complex numbers: <math>y</math> represents the complex number <math>6+i</math>, and <math>x+y=\frac{\pi}{4}</math> represents the fact <math>x\cdot y=1+i</math>, so dividing <math>1+i</math> by <math>6+i</math>, we get <math>\frac{7+5i}{56}</math>, which means <math>\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{iTest box|year=2007|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category:Introductory Trigonometry Problems]] | ||
Latest revision as of 20:30, 4 February 2023
Contents
Problem
If 
and 
are acute angles such that 
and 
, find the value of 
.
Solution
From the second equation, we get that 
. Plugging this into the first equation, we get:
Taking the tangent of both sides,
From the tangent addition formula, we then get:
.
Rearranging and solving, we get
Solution 2
We will use complex numbers: represents the complex number 
, and represents the fact 
, so dividing 
by , we get 
, which means
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 16 |
Followed by: Problem 18 | |
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