2007 iTest Problems/Problem 17: Difference between revisions
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== Solution == | == Solution == | ||
From the second equation, we get that <math>y=arctan | 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>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>\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1</math>. From the tangent addition formula, we then get: | ||
<math>\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}=1</math> | ||
<math>\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}</math>. Rearranging and solving, we get: | <math>\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}</math>. Rearranging and solving, we get: | ||
<math>\tan{x}=\box{\frac{5}{7}}</math> | <math>\tan{x}=\box{\frac{5}{7}}</math> | ||
Revision as of 05:25, 30 July 2016
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:
$x+arctan{\frac{1}{6}=\frac{\pi}{4}$ (Error compiling LaTeX. Unknown error_msg). Taking the tangent of both sides,
$\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1$ (Error compiling LaTeX. Unknown error_msg). From the tangent addition formula, we then get:
. Rearranging and solving, we get:
$\tan{x}=\box{\frac{5}{7}}$ (Error compiling LaTeX. Unknown error_msg)
