Doctor Q! (Math peoples please come)

[fireshot91]

My friend got these 2 problems to see if she can do it. I tried, my friends tried. Nobody can get it!!

[macjram]

I forgot all my geometry, and I just took it last year -_- lmao

[wywern209]

is the triangle in the first one normal( equilateral?)

[fireshot91]

Who knows! I thought that at first, and that would help with everything...but it doesn't specify.

[fireshot91]

What does that have to do with anything? What if the circles were bigger but the triangles position was the same..thus making it be equiangular.

[wywern209]

it has a lot to do with whether i can solve it or not. i could prob solve it right now but i need angles. lol, i feel extremely lazy right now.

[fireshot91]

Well yes, if it had angles, that basic geometry.

[sushi]

Both are trig problems.

Regarding the diagram, I say X is equal to 3.

[fireshot91]

How'd you get 3?

And yes, both are trig problems.

[wywern209]

if its basic trig, then its possible. with sine, tangent, and cosine.

[aethelbert]

The first one is

(16pi-6√12) / (16pi)

I think... I don't have a calculator to get the exact values, though.

[fireshot91]

Yes, I believe your answers but care to explain how you got them?

[mcavjame]

Area circle = Pi r2
=4.13 x 42
= 50.24 sq units

Area of triangle:
radius is 4 units
three triangles can be created using a ray extending from the center
the angle of each triangle at this center point must be 120o (360 / 3)
so now you have two sides and an enclosed angle.
use your cosine laws to calculate the area of one of these segments is 6.92
multiplied by 3 = 20.76

The shaded area is 50.24 - 20.76 = 29.48
Your chances are 29.48 in 50.24

[fireshot91]

Area circle = Pi r2
=4.13 x 42
= 66.08 sq units

Area of triangle:
radius is 4 units
three triangles can be created using a ray extending from the center
the angle of each triangle at this center point must be 120o (360 / 3)
so now you have two sides and an enclosed angle.
use your cosine laws to calculate the area of one of these segments is 6.92
multiplied by 3 = 20.76

The shaded area is 66.08 - 20.76 = 45.32
Your chances are 45.32 in 66.08



Isn't it 3.14 * 16 since pi- 3.14 and 16 is r squared.

[aethelbert]

I also got 3 for the second one.

[darklyt]

The first one is

(16pi-6√12) / (16pi)

I think... I don't have a calculator to get the exact values, though.

Yes, that's correct. It's a bit cleaner to write (4pi - 3*sqrt(3))/4pi. The answer you gave is exact, decimals are approximations.



Assuming that what you drew is an equilateral triangle in the circle, you have that 4 is the radius of the circle. Hence, 16pi is the area of the circle. You can form an isosceles triangle with legs of length 4 on each side, with the base the length of the side of the equilateral triangle. Dividing this into two 30-60-90 triangles, you find that the sides are 2, 2*sqrt(3), and 4. Thus the height of the triangle is 6 and the length of a side is 4*sqrt(3). The area is 12*sqrt(3). Therefore, the area of the black shaded area is 16pi - 12*sqrt(3) and the percentage of that area with respect to the total area of the circle is (16pi - 12*sqrt(3))/16pi = (4pi - 3*sqrt(3))/4pi.

[aethelbert]

Yes, I believe your answers but care to explain how you got them?
It's just trig... Every length there is the raduis, so inside the big triangle you can think of 3 smaller ones with two sides equal to four and two angles equal to 30 (it's equilateral, divide by 3 for 60, divide by two to split angles). It's a 30-60-90 triangle, so solve with the √3 stuff. You get √12 for the base and 2 for the other side when you divide the big thing into six. So .5(base)=√12 and your height is 4+2=6. That's the area of the triangle, 6√12. You know how to get the circle, so just find the difference and divide by the total area of the circle.

[mcavjame]

Isn't it 3.14 * 16 since pi- 3.14 and 16 is r squared.

yep... just caught that and recalculated... thanks. Also I was using r2 to mean r squared... no superscript here.

[aethelbert]

Yes, that's correct. It's a bit cleaner to write (4pi - 3*sqrt(3))/4pi.
Good thing I have no teacher anymore to take off points for not simplifying fractions!

[darklyt]

Good thing I have no teacher anymore to take off points for not simplifying fractions!

Or radicals lol.

[aethelbert]

How'd you get 3?
For the second one, the sides of the lower triangle are 6, 8, 10. Use inverse tangent to get the top right angle equal to about 36.8º. That means that it also applies to the bottom left angle on the top left triangle since the lines are parallel, making the angles basically vertical. You know the hypotenuse is 5, the other angles are 90 and 43.2. Use use the sine of 36.8 with 5 and you get 3.

Or radicals lol.
No radicals in the denominator, garbage.

[fireshot91]

Oh, I get it. Its the height of it, so it wouldn't matter if you move it left or right. The height will remain the height.

[sushi]

How'd you get 3?

And yes, both are trig problems.
The drawing is not to scale which makes it hard to see.

The top diagram is a possible parallelogram. We need to verify.

The square at the bottom means a right angle. Since the line with the 5 (and black triangle) is straight, the the opposite corner angles must be equal to 90 degrees. This makes the top object a parallelogram.

The top of the triangle, with 2 black triangles, is 10.

The left side line of the parallelogram, with one black triangle, is 5.

For a sanity check, the right side of the triangle is 8. So the line should be longer than the 6 one. Of course seeing the 6 side of the triangle and the 5 side of the parallelogram is a good indication of the distortion.

Anyhow, a quick arccos of (6/10) is 53.13 degrees.

To find the remaining angle, or angle between the left side of the parallelogram (one with only the black triangle) and the horizontal line we must subtract 53.13 degrees from 90 degrees which gives us 36.87 degrees.

To calculate X, I took the Sin (36.87) X 5 which is 3.

Oh, I get it. Its the height of it, so it wouldn't matter if you move it left or right. The height will remain the height.
Once we verified that the top is a parallelogram, then yes, the distance for X would remain constant across the parallelogram.

[darklyt]

For the second one, the sides of the lower triangle are 6, 8, 10. Use inverse tangent to get the top right angle equal to about 36.8º. That means that it also applies to the bottom left angle on the top left triangle since the lines are parallel, making the angles basically vertical. You know the hypotenuse is 5, the other angles are 90 and 43.2. Use use the sine of 36.8 with 5 and you get 3.

Nice solution.

No radicals in the denominator, damnit.

Nahh, that's for the old days, when you actually had to divide. Much easier to divide an integer into an irrational than an irrational into an integer lol. Thank god for calculators! No real need to do all that simplification stuff.

[aethelbert]

Nahh, that's for the old days, when you actually had to divide. Much easier to divide an integer into an irrational than an irrational into an integer lol. Thank god for calculators! No real need to do all that simplification stuff.
Indeed it was, haha. I'm impressed that I can still find inverse trig functions in on paper.

[Abstract]

Coming from someone who hasn't done trig in ages (since high school), I don't think these are very difficult.

I'm a bit against giving people answers for homework (although I don't mind giving tips), but the OP clearly has all the info he needs to do it himself now. ;)

[benbondu]

The first problem is kind of silly. I don't think there's enough information. I guess you could assume a uniform distribution inside of the circle and zero probability outside of the circle. If the dart has yet to be thrown, and you assume the person throwing it is aiming for the circle (presumably the center in order to minimize the chances of the dart landing outside), I find it hard to believe a uniform distribution is possible.

I might like the problem better if it said "given that a dart has landed in this circle, what is the probability it landed outside of the triangle?" That way you don't have to assume the thrower was aiming at anything in particular.

You could also assume the circle is the entire universe and the throw is completely random. I'd imagine the circle as the ground with walls extending upward. You throw something up in the air. It might bounce off the walls a few times, but it eventually has to land somewhere on the circle. But why use the image of a dart throw in that case? Overall, a poorly worded problem.

Or you could just cut the BS and ask "what percentage of this circle is shaded?"

The second problem doesn't require any trigonometry. Plane geometry suffices.

[Doctor Q]

This kind of thread is so much more interesting than threads about computers!

[swiftaw]

For the first one, assuming each point is equally likely for the dart to land, then the probability of landing in the shaded area = area of shaded part / total area

total area = 16*pi

area of shaded part = 16*pi - 12sqrt(3)

so probability = 1 - (3*sqrt(3) / 4*pi) = 0.5865

Where did these come from. Well, total area is easy, it's a circle so area is pi*r^2 = pi*4^2 = 16*pi

area of shaded part = area of circle - area of triangle.

Okay, how to find the area of the triangle. Divide up the big triangle into 3 identical isosceles triangles by drawing lines from the center to each vertex. Each of those isosceles triangles has two sides of length 4 and an angle of 120 degrees. Basic trig shows the area of each is 4*sqrt(3), so the big triangle has area 12*sqrt(3).

[JohnDoe8450]

The top diagram is a possible parallelogram. We need to verify.

Once we verified that the top is a parallelogram, then yes, the distance for X would remain constant across the parallelogram.

It is a parallelogram.
The little triangles mean that the lines are parallel.
Basically ----->---- is parallel to ----->----------
and ------>-->----- is parallel to -------->->------

At least, that's the notation we used...

[sushi]

It is a parallelogram.
Yes, I know that it is. If you read my post entirely you will see that I come to that conclusion.

Note, I may have not expressed my point well and for that I apologize.

Anyhow, an old adage for engineers, software developers and pilots, never assume! Always verify! :)

[eldy]

Area circle = Pi r2
=4.13 x 42
= 50.24 sq units

Area of triangle:
radius is 4 units
three triangles can be created using a ray extending from the center
the angle of each triangle at this center point must be 120o (360 / 3)
so now you have two sides and an enclosed angle.
use your cosine laws to calculate the area of one of these segments is 6.92
multiplied by 3 = 20.76

The shaded area is 50.24 - 20.76 = 29.48
Your chances are 29.48 in 50.24

You forgot the area of the dot, line and number four. :p