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Difference between revisions of "2015 AMC 12B Problems/Problem 13"

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==Problem==
 
==Problem==
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Quadrilateral <math>ABCD</math> is inscribed in a circle with <math>\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,</math> and <math>BC=6</math>. What is <math>AC</math>?
  
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<math>\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7</math>
  
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==Solution==
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<math>\angle ADB</math> and <math>\angle ACB</math> are both subtended by segment <math>AB</math>, hence <math>\angle ACB = \angle ADB = 40^\circ</math>. By considering <math>\triangle ABC</math>, it follows that <math>\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ</math>. Hence <math>\triangle ABC</math> is isosceles, and <math>AC = BC = \boxed{\textbf{(B)}\; 6}.</math>
  
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==Solution==
 
  
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Note: The statement <math>AD=4</math> is not needed.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=14|num-b=12}}
 
{{AMC12 box|year=2015|ab=B|num-a=14|num-b=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:52, 10 June 2024

Problem

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$. What is $AC$?

$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$

Solution

$\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$, hence $\angle ACB = \angle ADB = 40^\circ$. By considering $\triangle ABC$, it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$. Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{\textbf{(B)}\; 6}.$


Note: The statement $AD=4$ is not needed.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AMC 12 Problems and Solutions

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