Difference between revisions of "1987 AHSME Problems/Problem 6"
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\textbf{(D)}\ 2w-y-z\qquad | \textbf{(D)}\ 2w-y-z\qquad | ||
\textbf{(E)}\ 180-w+y+z</math> | \textbf{(E)}\ 180-w+y+z</math> | ||
| + | |||
| + | == Solution == | ||
| + | By angles in a quadrilateral, <math>x = 360^{\circ} - \text{reflex } \angle ADB - y - z</math>, and by angles at a point, <math>\text{reflex } \angle ADB = 360^{\circ} - w</math>, so our expression becomes <math>360^{\circ} - (360^{\circ} - w) - y - z = w - y - z</math>, which is <math>\boxed{\text{A}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 11:32, 1 March 2018
Problem
In the 
shown, 
is some interior point, and 
are the measures of angles
in degrees. Solve for 
in terms of 
and 
.
![[asy] draw((0,0)--(10,0)--(2,7)--cycle); draw((0,0)--(4,3)--(10,0)); label("A", (0,0), SW); label("B", (10,0), SE); label("C", (2,7), W); label("D", (4,3), N); label("x", (2.25,6)); label("y", (1.5,2), SW); label("$z$", (7.88,1.5)); label("w", (4,2.85), S); [/asy]](http://latex.artofproblemsolving.com/2/3/9/2399ef53c5ca3aa28698a8213a9aeb2fe7a397de.png)
Solution
By angles in a quadrilateral, 
, and by angles at a point, 
, so our expression becomes 
, which is 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
