2 humps on a Bactrian Camel.
I think C B stand for Cornerback.
A B C on the tree . A stand for A B stand for Bee C stand for see means " A bee see on the tree " .....
well it could be anything, but I think you are thinking of Pythagoras' theorem. a(2) x b(2) = c(2) (2) = squared
2 Humps on a Bactrian Camel (just one on a dromedary).
A,b,c
This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)
The Law of Sines: with triangle ABC, the angles are A, B, & C. The sides {a, b, & c} are opposite of the respective capital letter vertex.a/sin(A) = b/sin(B) = c/sin(C). You know the angles A, B, C; and two sides (say a & b).So side c = a*sin(C)/sin(A) = b*sin(C)/sin(B).You could also use the Law of Cosines: c^2 = a^2 + b^2 - 2*a*b*cos(C)
You must know the other side. It could be anything greater than zero yet less than 5 and still be a right triangle.a^2 +b^2=c^25^2 +b^2=c^225+b^2=c^225=c^2-b^2
before christ
If you know two sides of a right angle triangle, you can figure out the third by using the following formula: A*A+B*B=C*C or A**2+B**2=C**2 or C*C-B*B=A*A or C**2-B**2=A**2 or C*C-A*A=B*B or C**2-A**2=B**2 A is the bottom side if the right angle is at the bottom left, B is The only vertical(strait up and down) line if A is correct, and C is the last line.
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
Suppose sqrt(A) = B ie the square with sides B has an area of A and its perimeter is 4*B. Now consider a rectangle with sides C and D whose area is A. So C*D = A = B*B so that D = B*B/C Perimeter of the rectangle = 2*(C+D) = 2*C + 2*D = 2*C +2*B*B/C Now consider (C-B)2 which, because it is a square, is always >= 0 ie C*C + B*B - 2*B*C >= 0 ie C*C + B*B >= 2*B*C Multiply both sides by 2/C (which is >0 so the inequality remains the same) 2*C + 2*B*B/C >= 4*B But, as shown above, the left hand side is perimeter of the rectangle, while the right hand side is the perimeter of the square.