Difference between revisions of "1972 AHSME Problems/Problem 27"

(Created page with "==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</ma...")
 
 
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==Solution==
 
==Solution==
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Draw Diagram later
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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>.
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<math>\newline</math>
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Squaring both sides gives us that <math>rs=144</math>. We can substitute this value into the equation we had earlier. This gives us
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<math>\newline</math>
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<math>\frac{1}{2}\times144\times\sin A=64</math>
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<math>\newline</math>
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<math>72\sin A=64</math>
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<math>\newline</math>
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<math>\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}</math>

Latest revision as of 13:27, 13 January 2018

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 $\sqrt{rs}=12$. $\newline$ Squaring both sides gives us that $rs=144$. We can substitute this value into the equation we had earlier. This gives us $\newline$ $\frac{1}{2}\times144\times\sin A=64$ $\newline$ $72\sin A=64$ $\newline$ $\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}$

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