Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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− | + | ==Problem== | |
+ | In <math>\bigtriangleup ABC</math>, <math>D</math> is a point on side <math>\overline{AC}</math> such that <math>BD=DC</math> and <math>\angle BCD</math> measures <math>70^\circ</math>. What is the degree measure of <math>\angle ADB</math>? | ||
+ | |||
+ | <asy> | ||
+ | size(300); | ||
+ | defaultpen(linewidth(0.8)); | ||
+ | pair A=(-1,0),C=(1,0),B=dir(40),D=origin; | ||
+ | draw(A--B--C--A); | ||
+ | draw(D--B); | ||
+ | dot("$A$", A, SW); | ||
+ | dot("$B$", B, NE); | ||
+ | dot("$C$", C, SE); | ||
+ | dot("$D$", D, S); | ||
+ | label("$70^\circ$",C,2*dir(180-35));</asy> | ||
+ | |||
+ | <math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | ||
+ | ==Solution== | ||
+ | <math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>. Then <math>\angle CDB = 180-(70+70) = 40</math>. Since <math>\angle ADB</math> and <math>\angle BDC</math> are supplementary, <math>\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2014|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:16, 13 December 2019
Problem
In , is a point on side such that and measures . What is the degree measure of ?
Solution
, so . Then . Since and are supplementary, .
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.