A math teacher in the building I work in proposed this question to her 7th grade students, and it has me stumped. Any help would be appreciated.
You have a triangle with angle measures of 15, 45, and 120 degrees. The area of the triangle is equal to the quantity: m-n times the square root of g. Can’t get a square root symbol, but it’s the product of “n” and the square root of “g”–no parentheses used.
The students are supposed to give the value of m, n, and g.
I don’t have an answer, but just to clarify, since you said “no parentheses:”
area = m - n * g[sup]½[/sup] and not
area = (m - n) * g[sup]½[/sup]
:dubious: How do m, n, and g relate to the shape of the triangle?
The first example is correct.
There are infinitely many different triangles with angles of 15, 45, and 120 degrees, and each has a different area. Were you given any length, or is the solution supposed to be in terms of the lengths of one or more of the sides?
Are we supposed to assume Euclidean geometry? 
The only information given is that the expression given above is equal to the area of the triangle. In my limited math knowledge, I see no way that this is possible to solve.
Ok…So apparently the students who came to see me left out one small detail: the side opposite of the 45 degree is equal to 20 units.
I can now solve for the area of the triangle using some trig., but I still get stuck on how to figure the invidivual values of the three values (m, n, g).
Taking this at face value for a minute, the answer must be yes, or else the angles wouldn’t add to 180.
I now return to the joke, already in progress.
Piece of cake. See http://people.bath.ac.uk/masgcs/Article40.pdf
Do they know the basic trigonometric relations for 45-45-90 and 30-60-90 triangles? If so, here’s how I’d do it:
Draw the triangle with the base being the side connecting the 45- and 120-degree angles, and the 15-degree angle to the upper right. Then you can drop a perpendicular from the 15-degree vertex down to the baseline, and extend the base out to the perpendicular you just drew. If you look at that for a second, you’ll see that you’ve now got a 45-45-90 triangle that’s constructed from the original one plus a new 30-60-90 triangle. And so the area of the original is the difference in the areas of those two triangles. Now note that the hypotenuse of the 45-45-90 is the length of the side of the original that is opposite the 120-degree angle. You can now express everything in terms of that length by using what you know about the side length ratios of those particular right triangles.
Hope that helps.
I assume m, n, and g must all be integers?
Given your additional info that the side opposite the 45 degree angle is 20 units, I get that the area is:
150 - 50 * 3[sup]1/2[/sup]
Which just highlights another issue with the problem as stated. The values m, n, and g, even if assumed to be integral, are not unique, since your answer of m = 150, n = 50, and q = 3 is equivalent to the answer I reached with n = 5 and q = 300. Presumably the teacher wants your answer, however.
Assuming that students have seen 30-60-90 and 45-45-90 triangles and that the teacher posed the problem well, I like it. It should be within reach of any student who has a grasp on the prerequisite material, and it’s definitely not routine.
I don’t know of a solution that 7th graders are expected to know, but here’s what I got using what is normally 11th-grade math:
Let S be the length of the side opposite the 120-degree angle.
By the Law of Sines, sin 45 / 20 = sin 120 / S -> S = 20 sin 120 / sin 45 = 10 sqrt(6).
Let A, B, and C be the vertices of the triangle with angles 120, 45, and 75 respectively.
Construct the perpendicular to BC that goes through A; label the intersection point D.
Let H be the length of AD, which is the “height” of the triangle with base S.
ABD is a 45-45-90 triangle, so DB = AD = h, and AB = sqrt(2) h.
Using the “sine of differences” formula, sin 15 = sin (45 - 30) = sin 45 cos 30 - cos 45 sin 30 = (sqrt(2)/2 x sqrt(3)/2) - (sqrt(2)/2 - 1/2) = (sqrt(6) - sqrt(2)) / 4.
Again using the Law of Sines, sin 45 / 20 = sin 15 / (h sqrt(2)) -> h = (20 sin 15) / (sqrt(2) sin 45) = (20 x (sqrt(6) - sqrt(2) / 4) / (sqrt(2)/2 x sqrt(2)) = 5 (sqrt(6) - sqrt(2).
The area of the triangle = 1/2 x S x H = 1/2 x 10 sqrt(6) x 5 (sqrt(6) - sqrt(2))
= 25 sqrt 6 (sqrt(6) - sqrt(2))
= 25 x 6 - 25 x sqrt(12)
= 150 - 25 x (2 sqrt(3))
= 150 - 50 sqrt(3).
m = 150, n = 100, and g = 3.
This is what I needed to see. Thanks!
I like the question as well, and the students do know the triangle relationships. I definitely don’t see any of them solving this one on their own though. Hope I’m wrong.
Yeah, by force of long, long habit, I always make the implicit assumption that answers to problems of this type should be in “normal form”, where all roots are removed from denominators, and any factors of a radicand that can be pulled out of the radical, must be pulled out.
But I’m seriously anal-retentive about stuff like this.
OOPS - that should be n = 50.
Of course, RickG’s solution is far more in the realm of what 7th graders might know.
I don’t worry about it (usually) for my students, but that’s only because at the college level, you tend to assume (usually justifiably) that students already know how to work with roots and simplify.
That Don Guy
That’s what it says in the link I posted.