Difference between revisions of "1961 AHSME Problems/Problem 8"

Latest revision as of 14:36, 19 May 2018

Problem

Let the two base angles of a triangle be $A$ and $B$ , with $B$ larger than $A$ . The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$ , with $C_2$ adjacent to side $a$ . Then:

$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$

Solution

$[asy] draw((0,0)--(120,0)--(40,50)--(0,0)); draw((40,50)--(40,0)); draw((40,4)--(44,4)--(44,0)); label("$B$",(10,5)); label("$A$",(100,5)); label("$C_1$",(47,38)); label("$C_2$",(34,33)); [/asy]$

Noting that side $a$ is opposite of angle $A$ , label the diagram as shown.

From the two right triangles, $B + C_2 = 90$ $A + C_1 = 90$ Substitute to get $B + C_2 = A + C_1$ $B - A = C_1 - C_2$ The answer is $\boxed{\textbf{(B)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 33: / Line 33: @@
 ==See Also==
-{{AHSME 40p box|year=1961|num-b=2|num-a=4}}
+{{AHSME 40p box|year=1961|num-b=7|num-a=9}}
 {{MAA Notice}}
 [[Category:Introductory Geometry Problems]]

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Difference between revisions of "1961 AHSME Problems/Problem 8"

Latest revision as of 14:36, 19 May 2018

Problem

Solution

See Also