2018 AMC 12B Problems/Problem 23

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 $\angle ACB$?

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

Solution

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 $\boxed{\textbf{C}.}$


Note (not by author): Alternatively, to find the angle without dot products, one may compute the distance from $A$ to $B$ as

$\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}=\sqrt{3}$.

From the Law of Cosines, $3=1^2+1^2-2\cos{\theta}$, so $\cos{\theta}=-\frac{1}{2},$ from which the desired conclusion follows.

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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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