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1972 AHSME Problems/Problem 27: Difference between revisions

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<math>\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}</math>
<math>\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}</math>
== Problem ==
If the area of <math>\triangle ABC</math> is <math>64</math> square units and the geometric mean (mean proportional)
between sides <math>AB</math> and <math>AC</math> is <math>12</math> inches, then <math>\sin A</math> is equal to
<math>\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad
\textbf{(B) }\frac{3}{5}\qquad
\textbf{(C) }\frac{4}{5}\qquad
\textbf{(D) }\frac{8}{9}\qquad
\textbf{(E) }\frac{15}{17} </math> 


==Solution==
==Solution==
Line 9: Line 20:


We can let <math>AB=s</math> and <math>AC=r</math>. We can also say that the area of a triangle is <math>\frac{1}{2}rs\sin A</math>, which we know, is <math>64</math>. We also know that the geometric mean of <math>r</math> and <math>s</math> is 12, so <math>\sqrt{rs}=12</math>.
We can let <math>AB=s</math> and <math>AC=r</math>. We can also say that the area of a triangle is <math>\frac{1}{2}rs\sin A</math>, which we know, is <math>64</math>. We also know that the geometric mean of <math>r</math> and <math>s</math> is 12, so <math>\sqrt{rs}=12</math>.
<math>\newline</math>
 
 
Squaring both sides gives us that <math>rs=144</math>. We can substitute this value into the equation we had earlier. This gives us  
Squaring both sides gives us that <math>rs=144</math>. We can substitute this value into the equation we had earlier. This gives us  
<math>\newline</math>
<cmath>\frac{1}{2}\times144\times\sin A=64</cmath>
<math>\frac{1}{2}\times144\times\sin A=64</math>
<cmath>72\sin A=64</cmath>
<math>\newline</math>
<cmath>\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}</cmath>
<math>72\sin A=64</math>
 
<math>\newline</math>
-edited for readability
<math>\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}</math>

Revision as of 09:39, 29 January 2021

Problem

If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to

$\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}$

Problem

If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to

$\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}$

Solution

Draw Diagram later

We can let $AB=s$ and $AC=r$. We can also say that the area of a triangle is $\frac{1}{2}rs\sin A$, which we know, is $64$. We also know that the geometric mean of $r$ and $s$ is 12, so $\sqrt{rs}=12$.


Squaring both sides gives us that $rs=144$. We can substitute this value into the equation we had earlier. This gives us \[\frac{1}{2}\times144\times\sin A=64\] \[72\sin A=64\] \[\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}\]

-edited for readability

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