Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 28"

 
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== Problem ==
 
== Problem ==
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Suppose <math>\triangle ABC</math> is a triangle with 3 acute angles <math>A, B,</math> and <math>C</math>.  Then the point <math>( \cos B - \sin A, \sin B - \cos A)</math>
  
<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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(A) can be in the 1st quadrant and can be in the 2nd quadrant only
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(B) can be in the 3rd quadrant and can be in the 4th quadrant only
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(C) can be in the 2nd quadrant and can be in the 3rd quadrant only
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(D) can be in the 2nd quadrant only
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(E) can be in any of the 4 quadrants
  
 
== Solution ==
 
== Solution ==
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First note that <math>\cos{A}, \cos{B}, \sin{A}, \sin{B}>0</math> and <math>A+B>90</math>. From these, we deduce that the <math>x</math> value and the <math>y</math> value of the coordinates will always be positive and thus the point can only be in the first quadrant.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]

Revision as of 19:23, 22 July 2006

Problem

Suppose $\triangle ABC$ is a triangle with 3 acute angles $A, B,$ and $C$. Then the point $( \cos B - \sin A, \sin B - \cos A)$

(A) can be in the 1st quadrant and can be in the 2nd quadrant only

(B) can be in the 3rd quadrant and can be in the 4th quadrant only

(C) can be in the 2nd quadrant and can be in the 3rd quadrant only

(D) can be in the 2nd quadrant only

(E) can be in any of the 4 quadrants

Solution

First note that $\cos{A}, \cos{B}, \sin{A}, \sin{B}>0$ and $A+B>90$. From these, we deduce that the $x$ value and the $y$ value of the coordinates will always be positive and thus the point can only be in the first quadrant.

See also

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