AC = sqrt(AB^2+BC^2) other wise known as a^2+b^2=c^2. Therefore AC is around 51.739
All the trigonometric functions are derived from the right angled triangle. If we consider the three sides (AB, BC, CA) of a triangle and the included angle. There is a possibility of getting six functions based on the ratios like AB/AC, BC/AC, AB/BC, BC/AB, AC/BC, AC/AB . So we will have six trigonometric functions
AB + AC + BC = 48 AB + (AB +9) + (AB + 9 + 3) = 48 Solve and AB = 9 So AB = 9, AC = 18 and BC = 21
To find the length of side AC in a triangle, you can use the Law of Cosines if you know the lengths of the other two sides (AB and BC) and the included angle (∠B). The formula is: [ AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle B) ] After calculating AC², take the square root to find AC. If you have a right triangle, you can simply use the Pythagorean theorem: [ AC = \sqrt{AB^2 + BC^2} ] (assuming AC is the hypotenuse).
Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
To find the possible length for side AB in triangle ABC with sides BC = 12 and AC = 21, we can use the triangle inequality theorem. The sum of the lengths of any two sides must be greater than the length of the third side. Therefore, we can write the inequalities: AB + BC > AC → AB + 12 > 21 → AB > 9 AB + AC > BC → AB + 21 > 12 → AB > -9 (which is always true) BC + AC > AB → 12 + 21 > AB → 33 > AB or AB < 33 Combining these, we get the inequality: 9 < AB < 33.
Assuming that AB and AC are straight lines, the answer depends on the angle between AB and AC. Depending on that, BC can have any value in the range (22, 46).
yes because ab plus bc is ac
All the trigonometric functions are derived from the right angled triangle. If we consider the three sides (AB, BC, CA) of a triangle and the included angle. There is a possibility of getting six functions based on the ratios like AB/AC, BC/AC, AB/BC, BC/AB, AC/BC, AC/AB . So we will have six trigonometric functions
AC=5 AB=8 A=1 B=8 C=5 BC=40
C is the midpoint of Ab . then AC = BC. So AC= CB.
AB + AC + BC = 48 AB + (AB +9) + (AB + 9 + 3) = 48 Solve and AB = 9 So AB = 9, AC = 18 and BC = 21
Ab+bc=ac
Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
there will be three vertex AB, BC, AC
ABC
It could be a vector sum.
There are some missing terms. First of all, I assume that A, B, and C are collinear and that B is between A and C.If this is true then AC-AB=BC by the whole is the sum of its parts theorem.24-20=4Otherwise, all that can be said about BC is that it's length is between AC-AB = 4 and AC+AB = 44 units.