I wish i was that smart !
I mean, Pythagorean theorem (and sin, tan, cos etc.) is the only way I can see it be done. I haven't had coffee today though so I might be temporarily retarded.
Just count the squares?
3 squares horizontal
7 squares vertical
h = sqrt(0,3² + 0,7²)
What's the entire question? The triangle is inscribed in a circle. You can find the angles using the vertical line as a reference to create 2 triangles, so you know you have two 45 degree angles and a 90 degree angle. You can also just use the grid in the back to know how many units long the line is. If it doesn't give you any units or any measurements to work with though you can't find it's length because it could just be dilated
Use trigonometry. And sin(A/2) = sqrt((1 - cos(A))/2)
Just so I'm understanding correctly, you want to calculate the chord length without using trigonometry?
find cords and then count the length ez
(BB`)^2=(AB)^2+(AB`)^2-2(AB)(AB`) x cos A
idk man i failed my geometry and trigonometry shit, i now understand a bit but thats some new high level shit
Anyway, the triangle made by A, B1 and that unspecified point is an isosceles triangle, since y = 90 and A = 45, so B1 is 180-90-45 = 45. Just remember this for any time you spot a right triangle with one angle being 46. So the short, equal sides are of some length, x, and by the Pythagorean theorem, the hypotenuse, which is 1, gives the equation x^2 + x^2 = 1^2, in other words, x = 0.5sqrt(2). AB1 = AB, since A is the middlepoint of a circle and B1 and B are both on it, so j = x = 0.5sqrt(2) and that short length on AB is 1- 0.5sqrt(2). Now use the Pythagorean theorem again.
h^2 = (0.5sqrt(2))^3 + (1-0.5sqrt(2))^2 and you can simplify that.
just use this formula men
chordlength = 2* radius*sin (A/2)
i got 0.765 units
Bro how can ı open fucking imgur sites
Just help yourself with the squares, use pythagorean theorem. I'm sure it would be pretty easy. Do you know the length of a square or something? If so then you can easily solve it.
If 1 is the length of |AB| then then one square is 0,1 and a square inside that square would be 0,02. That's all the info you need to solve it.
So easy that I can't stop laughing at retards on hltv.
Have you heard about something know as similarity and pythagoras theorem?
Let the foot of perpendicular be P
Then triangle AB'P is similar to triangle B'BP
So j^2=x*(1-x) where x is length of AP.
Now find value of j^2 from 2 right triangles and this value.
Equate all 3, factorise them and done.
is point a the center of the circle slice?
pythagoras in left triangle a^2 + a^2 = 1
j = a;
in right triangle base = 1-a;
now h^2 = j^2 + (1-a)^2;
if my calculations are correct the answer is 7.
Use Pythagorean to find bigger length of AB, once you know that then the short side of AB is just 1 - bigger side. Now you have the small side of AB and you can use pyth theorem to find h