Difference between revisions of "2021 JMC 10 Problems/Problem 13"
Skyscraper (talk | contribs) (Created page with "==Problem== An angle chosen from <math>1^{\circ},2^{\circ}, \dots,90^{\circ}</math> and an angle chosen from <math>1^{\circ},2^{\circ},\dots,89^{\circ}</math> determine two a...") |
Skyscraper (talk | contribs) (→Solution) |
||
Line 9: | Line 9: | ||
− | Suppose <math>A = n^{\circ}.</math> Then, <math>B</math> can equal <math>1^{\circ}, 2^{\circ} , \dots (90 - n - 1)^{\circ}.</math> We can see that there are <math>1+2+ \dots + 88=\tfrac{88\cdot 89}{2} = 44\cdot 89</math> desired cases and <math>89 \cdot 90</math> total cases, so the answer is < | + | Suppose <math>A = n^{\circ}.</math> Then, <math>B</math> can equal <math>1^{\circ}, 2^{\circ} , \dots (90 - n - 1)^{\circ}.</math> We can see that there are <math>1+2+ \dots + 88=\tfrac{88\cdot 89}{2} = 44\cdot 89</math> desired cases and <math>89 \cdot 90</math> total cases, so the answer is <cmath>\frac{44\cdot 89}{89 \cdot 90} = \frac{22}{45}.</cmath> |
Latest revision as of 16:16, 1 April 2021
Problem
An angle chosen from and an angle chosen from
determine two angles of a triangle. What is the probability this triangle is obtuse?
Solution
Let be the angle chosen from
and
be the angle chosen from
For the triangle to be obtuse, we must have
Suppose Then,
can equal
We can see that there are
desired cases and
total cases, so the answer is