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

Latest revision as of 12:49, 23 June 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

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

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

@@ Line 3: / Line 3: @@
 <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==

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

Latest revision as of 12:49, 23 June 2021

Problem

Solution