2021 JMC 10 Problems/Problem 13

Problem

An angle chosen from $1^{\circ},2^{\circ}, \dots,90^{\circ}$ and an angle chosen from $1^{\circ},2^{\circ},\dots,89^{\circ}$ determine two angles of a triangle. What is the probability this triangle is obtuse?

$\textbf{(A) } \dfrac{11}{45} \qquad\textbf{(B) } \dfrac{22}{45} \qquad\textbf{(C) } \dfrac{1}{2} \qquad\textbf{(D) } \dfrac{5}{9} \qquad\textbf{(E) } \dfrac{11}{15}$

Solution

Let $A$ be the angle chosen from $1^{\circ},2^{\circ}, \dots, 90^{\circ}$ and $B$ be the angle chosen from $1^{\circ},2^{\circ}, \dots, 89^{\circ}.$ For the triangle to be obtuse, we must have $180^{\circ}- A - B > 90^{\circ} \implies A + B < 90^{\circ}.$

Suppose $A = n^{\circ}.$ Then, $B$ can equal $1^{\circ}, 2^{\circ} , \dots (90 - n - 1)^{\circ}.$ We can see that there are $1+2+ \dots + 88=\tfrac{88\cdot 89}{2} = 44\cdot 89$ desired cases and $89 \cdot 90$ total cases, so the answer is $\frac{44\cdot 89}{89 \cdot 90} = \frac{22}{45}.$

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2021 JMC 10 Problems/Problem 13

Problem

Solution