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

Latest revision as of 22:23, 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:

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

Let $AB$ and $CD$ intersect at $E$ . We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$ .

Since $\sin BCD = \frac{3}{5}$ , we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$ . Thus, $EB=3$ and $EC=4$ from simple sin application.

$AD$ is the hypotenuse of right $\triangle AED$ , with leg lengths $AB+BE=7$ and $EC+CD=24$ . Thus, $AD=\boxed{\textbf{(E)}25}$

-WannabeCharmander

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "1979 AHSME Problems/Problem 24"

Latest revision as of 22:23, 21 June 2018

Problem 24

Solution

See also

@@ Line 12: / Line 12: @@
 ==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);
+//diagram by WannabeCharmander
+</asy>
+Let <math>AB</math> and <math>CD</math> intersect at <math>E</math>. We know that <math>\angle AED=90^{\circ}</math> because <math>\angle E = 180 - (A+D)=180-90 = 90^{\circ}</math>.
+Since <math>\sin BCD = \frac{3}{5}</math>, we get <math>\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}</math>. Thus, <math>EB=3</math> and <math>EC=4</math> from simple sin application.
+<math>AD</math> is the hypotenuse of right <math>\triangle AED</math>, with leg lengths <math>AB+BE=7</math> and <math>EC+CD=24</math>. Thus, <math>AD=\boxed{\textbf{(E)}25}</math>
+-WannabeCharmander
 == See also ==