Difference between revisions of "2018 AMC 12B Problems/Problem 23"

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== Problem ==
 
== Problem ==
  
Ajay is standing at point <math>A</math> near Pontianak, Indonesia, <math>0^\circ</math> latitude and <math>110^\circ \text{ E}</math> longitude. Billy is standing at point <math>B</math> near Big Baldy Mountain, Idaho, USA, <math>45^\circ \text{ N}</math> latitude and <math>115^\circ \text{ W}</math> longitude. Assume that Earth is a perfect sphere with center <math>C</math>. What is the degree measure of <math>\angle ACB</math>?
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Ajay is standing at point <math>A</math> near Pontianak, Indonesia, <math>0^\circ</math> latitude and <math>110^\circ \text{ E}</math> longitude. Billy is standing at point <math>B</math> near Big Baldy Mountain, Idaho, USA, <math>45^\circ \text{ N}</math> latitude and <math>115^\circ \text{ W}</math> longitude. Assume that Earth is a perfect sphere with center <math>C.</math> What is the degree measure of <math>\angle ACB?</math>
  
<cmath>\textbf{(A) }105 \qquad
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<math>\textbf{(A) }105 \qquad
 
\textbf{(B) }112\frac{1}{2} \qquad
 
\textbf{(B) }112\frac{1}{2} \qquad
 
\textbf{(C) }120 \qquad
 
\textbf{(C) }120 \qquad
 
\textbf{(D) }135 \qquad
 
\textbf{(D) }135 \qquad
\textbf{(E) }150 \qquad</cmath>
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\textbf{(E) }150 \qquad</math>
  
== Solution ==
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== Diagram ==
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<b>IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...</b>
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== Solution 1 (Law of Cosines) ==
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<b>IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...</b>
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== Solution 2 (Coordinate Geometry) ==
  
 
Suppose that Earth is a unit sphere with center <math>(0,0,0).</math> We can let
 
Suppose that Earth is a unit sphere with center <math>(0,0,0).</math> We can let
 
<cmath>A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).</cmath>The angle <math>\theta</math> between these two vectors satisfies <math>\cos\theta=A\cdot B=-\frac{1}{2},</math> yielding <math>\theta=120^{\circ},</math> or <math>\boxed{\textbf{C}}.</math>
 
<cmath>A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).</cmath>The angle <math>\theta</math> between these two vectors satisfies <math>\cos\theta=A\cdot B=-\frac{1}{2},</math> yielding <math>\theta=120^{\circ},</math> or <math>\boxed{\textbf{C}}.</math>
  
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<b>IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...</b>
  
Note (not by author): Alternatively, to find the angle without dot products, one may compute the distance from <math>A</math> to <math>B</math> as
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== Solution 3 (Coordinate Geometry) ==
 
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<b>IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...</b>
<math>\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}=\sqrt{3}</math>.
 
 
 
From the Law of Cosines, <math>3=1^2+1^2-2\cos{\theta}</math>, so <math>\cos{\theta}=-\frac{1}{2},</math> from which the desired conclusion follows.
 
  
 
==See Also==
 
==See Also==

Revision as of 15:38, 28 October 2021

Problem

Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$

$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$

Diagram

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

Solution 1 (Law of Cosines)

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

Solution 2 (Coordinate Geometry)

Suppose that Earth is a unit sphere with center $(0,0,0).$ We can let \[A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).\]The angle $\theta$ between these two vectors satisfies $\cos\theta=A\cdot B=-\frac{1}{2},$ yielding $\theta=120^{\circ},$ or $\boxed{\textbf{C}}.$

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

Solution 3 (Coordinate Geometry)

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions

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