Question 1203593
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An angle is obtuse when the angle is between 90 degrees and 180 degrees excluding each endpoint.
90 < angle < 180


Quadruple all sides of that inequality to get 360 < 4*angle < 720
If a quadrilateral could have all 4 obtuse angles, then the angles would sum to a value anywhere between 360 and 720, excluding those endpoints.


But recall that the sum of the interior angles of any quadrilateral is always 360 degrees. 
It comes from this formula:
S = 180(n-2)
where,
S = sum of interior angles
n = number of sides
When n = 4 it leads to S = 360
We can see that 360 < 4*angle < 720 wouldn't be possible due to S = 360.
This contradiction would allow us to rule out choice A.
Quadrilaterals cannot have all angles that are obtuse.
The most obtuse angles a quadrilateral can have is 3.


Choice B is ruled out because all four angles of any rectangle are always 90 degrees.


Use the formula
S = 180(n-2)
to plug in n = 3 and you should get S = 180
The sum of all three interior angles of any triangle is always 180 degrees.
Start from 90 < angle < 180 and triple each side to get 270 < 3*angle < 540
Unfortunately 180 is not in this interval, so we cannot have a triangle with all angles obtuse.
The most obtuse angles a triangle can have is 1.
If one angle was obtuse, then the other two angles must be acute. The other two acute angles must be in the interval 0 < angle < 90.
So we can rule out choice C as well.


If n = 6, then S = 180*(n-2) = 180*(6-2) = 720 is the sum of all six angles of a hexagon.
Also if all 6 angles of a hexagon are obtuse, then 90 < angle < 180 leads to 540 < 6*angle < 1080
The 720 is indeed between 540 and 1080, which means it is possible to have a hexagon with all obtuse angles.
In fact, any regular hexagon will have each interior angle of 720/6 = 120 degrees



Bonus Questions:<ol><li>What is the smallest positive integer value of n such that 90 < 180(n-2)/n < 180 is true?</li><li>Can a quadrilateral have exactly 3 right angles, where the 4th angle is not 90 degrees?</li></ol>
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