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Doctor Q! (Math peoples please come)
- Thread starter fireshot91
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Who knows! I thought that at first, and that would help with everything...but it doesn't specify.
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.
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.
The first one is
(16pi-6√12) / (16pi)
I think... I don't have a calculator to get the exact values, though.
(16pi-6√12) / (16pi)
I think... I don't have a calculator to get the exact values, though.
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
=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
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.
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.
yep... just caught that and recalculated... thanks. Also I was using r2 to mean r squared... no superscript here.
Good thing I have no teacher anymore to take off points for not simplifying fractions!
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.
No radicals in the denominator, garbage.
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.
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.
Once we verified that the top is a parallelogram, then yes, the distance for X would remain constant across the parallelogram.
Nice solution.
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.