I was going through some study materials and came across a problem where the answer suggested to use a formula I don't ever remember encountering.?ÿ It seemed simpler to break it down from scratch than to memorize their formula. So have you ever used it?
a = sqrt [2 * area * sin A/(sin B * sin C) ]
Here's a similar problem.?ÿ Would you have used their formula or found another way?
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I tend to just commit the law of sines and law of cosines to memory, manipulate and derive the rest.
A very handy realignment of the basic formula.
Not a formula that I know, so I would do something like this:
h = q came quickly, tan (63 26 06) from calculator knowing that I needed the tangent to relate p and q. Then you gotta know that an acre is 66*660 square feet. (I can remember chains, but not acres.)
A little bit of substitution, some calculator work, and the answer falls out.
I don't have much room left in my head for formulas, so it takes me a bit longer, but I find a lot of cool stuff along the way.
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@mathteacher Wow.?ÿ Great approach.
Interesting problem
The size of a triangle is not fixed by the three angles. Usually require at least one side length for a solution.
The area is an interesting way to fix the size.
I have a vague recollection of the area formula but certainly recall sin, cos and tan formulae.
Last time I checked, and you understand new technology can change anything, the area of an acre was 43560 square feet.?ÿ Of course, in Texas acres are much smaller so the locals can claim to own more of them.
I can remember little oddities like a rectangle that is 1 chain by 10 chains is an acre. So 66 ft times 660 ft is an acre, 43,560 square feet.
Now I can remember pi equals 3.14159 which is more digits than 43,560, but that's different.
Another rarely used but common formula is for the solution of a quadratic equation when b?ý is much greater than 4ac:
x=2c/[-b?ñsqrt(b?ý-4ac)]
That's pretty much the way I'd work it, but instead of doing it in square feet, if you express the area as 75 sq chains it works out nicely and the only thing you need the calculator (or tables) for is the tangent.
Another rarely used but common formula is for the solution of a quadratic equation when b?ý is much greater than 4ac:
x=2c/[-b?ñsqrt(b?ý-4ac)]
I'd never seen this form. After playing around with it I think it is fully equivalent to the usual form, and not just under the stated condition. But where is the advantage in it? Seems like just as much computation.
@bill93?ÿ
It has a problem when c = 0. Then (-b + sqrt(b)) in the denominator is 0, 2c in the numerator is 0, and the fraction is indeterminate.
It has a problem when c = 0.?ÿ
Thus likely computational inaccuracy when ac is small. If c=0 we wouldn't usually be using the formula. It factors easily then.
@bill93?ÿ
Yes we would, and, conversely, when c is very small, we could ignore it (set it equal to 0) and solve by factoring. Consider x^2 + 6x - 0.001 = 0.
It's a great piece of mathematics, though, and thanks to @dave-lindell for sharing it.
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@aliquot?ÿ
I think it's more a question of accuracy. In the example x^2 + 6x -0.001 = 0,?ÿ the actual roots to 7 places are 0.0001667 and -6.0001667.
Ignoring the constant term and factoring yields 0.0000000 and -6.0000000. Ignoring the constant term implicitly assumes that one solution is zero.?ÿ
Of course, in today's world of automation, such simplifications are not really necessary, but they do offer a reasonableness check on computer solutions.