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Difference between revisions of "1979 AHSME Problems/Problem 24"

m (→‎Solution)
(→‎Solution)
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==Solution==
 
==Solution==
 +
We know that <math>\sin(C)=-\cos(B)=\frac{3}{5}</math>. Since <math>B</math> and <math>C</math> are obtuse, we have <math>\sin(180-C)=\cos(180-B)=\frac{3}{5}</math>. It is known that <math>\sin(x)=\cos(90-x)</math>, so <math>180-C=90-(180-C)=180-B</math>. We simplify this as follows:
 +
 +
<cmath>-90+C=180-B</cmath>
 +
 +
<cmath>B+C=270^{\circ}</cmath>
 +
 +
Since <math>B+C=270^{\circ}</math>, we know that <math>A+D=360-(B+C)=90^{\circ}</math>. Now extend <math>AB</math> and <math>CD</math> as follows:
 +
 +
<asy>
 +
size(10cm);
 +
label("A",(-1,0));
 +
dot((0,0));
 +
label("B",(-1,4));
 +
dot((0,4));
 +
label("E",(-1,7));
 +
dot((0,7));
 +
label("C",(4,8));
 +
dot((4,7));
 +
label("D",(24,8));
 +
dot((24,7));
 +
draw((0,0)--(0,4));
 +
draw((0,4)--(4,7));
 +
draw((4,7)--(24,7));
 +
draw((24,7)--(0,0));
 +
draw((0,4)--(0,7), dashed);
 +
draw((0,7)--(4,7), dashed);
 +
</asy>
 +
 +
Let <math>AB</math> and <math>CD</math> intersect at <math>E</math>.
 +
 
<math>\boxed{\textbf{E}}</math>
 
<math>\boxed{\textbf{E}}</math>
  

Revision as of 23:18, 21 June 2018

Problem 24

Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length

  • A polygon is called “simple” if it is not self intersecting.

$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$

Solution

We know that $\sin(C)=-\cos(B)=\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$. It is known that $\sin(x)=\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:

\[-90+C=180-B\]

\[B+C=270^{\circ}\]

Since $B+C=270^{\circ}$, we know that $A+D=360-(B+C)=90^{\circ}$. Now extend $AB$ and $CD$ as follows:

[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); [/asy]

Let $AB$ and $CD$ intersect at $E$.

$\boxed{\textbf{E}}$

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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