Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | <math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | ||
==Solution== | ==Solution== | ||
− | BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = <math>\boxed{140}</math> | + | BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = <math>\boxed{140}</math>, or <math>\boxed{D}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=8|num-a=10}} | {{AMC8 box|year=2014|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:43, 3 November 2015
Problem
In ,
is a point on side
such that
and
measures
. What is the degree measure of
?
Solution
BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = , or
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 |
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