09A Test

[calcgal]

A triangle has a fixed side dimension of length 7 in, and the opposite angle is also fixed and equal to 37 degrees. What is the maximum triangle area? (The answer is 36.6, but I can't figure out how they get it!)

[sinkokuyou]

set x as the side between 37deg and 7
and [unparseable or potentially dangerous latex formula] to be the angle opposite of x
[unparseable or potentially dangerous latex formula]
so [unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
draw it on your calculator, then you'll find the max=36.6

[austorior]

I just thought since one angle is 37, the other two angles add up to 143. To get the maximum area the two angles has to be as close as possible. The closest possible is 143/2 which is 71.5. You know the 3 angles and one side, get the three sides, and find the area with heron's or [unparseable or potentially dangerous latex formula]. By the way, even if you do the three angles as 37,71,72, the answer is still the same. Maybe this is the same explanation as the one above me but this may clear up any confusions if you have any...

[88bobcat]

[AuSmith]

If you recall a fact from geometry class, "if an angle with measure [unparseable or potentially dangerous latex formula] is inscribed in a circle, it cuts an arc with measure [unparseable or potentially dangerous latex formula]"

So, you can rephrase this problem as "A circle has a chord of length [unparseable or potentially dangerous latex formula], which subtends an inscribed angle of [unparseable or potentially dangerous latex formula]. What is the maximum area of the triangle formed by the chord and the angle?"

Typical triangle.

09A Inscribed Angle.jpg (6.77 KiB) Viewed 1175 times

But, I see no good solution from this... to make some sense of why the triangle must be isosceles, I say if I jog the [unparseable or potentially dangerous latex formula] vertex ever so slightly closer to the middle, the triangle gains area on the blue side and loses area on the red side. Since the blue is larger than the red, it should make sense that the maximum occurs when the red length equals the blue length.

Surely, this argument can be made precise via some "little-oh" notation and right triangles.