Angle abc.
The definition of a triangle is the union of segments AB, BC, and AC (where A, B, and C are not all collinear).
AC = 2*sqrt(3) = 3.4641AB = 2*sqrt(7) = 5.2915 angle ACB = 90 degrees.
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).
Draw two lines AB and AC that meet at point A. The angle BAC is greater than 90° but less than 180°. Let AB > AC. Draw a third line BC to complete the triangle so that BC is not equal to AB or AC. The triangle is a scalene triangle containing an obtuse angle.
Angle abc.
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
Angle B and Angle C
Yes. AB, AC, BC and EF.
Consider a right triangle ABC as shown below. The right angle is at B, meaning angle ABC is 90 degrees. With the editor I have, I am not able to draw the line AC but imagine it to be there. By pythagorean theorem AC*2 = AB*2 + BC*2. The line AC is called the hypotenuse. Consider the angle ACB. The cosine of this angle is BC/AC, the sine is AB/AC and tangent is AB/BC. If you consider the angle BAC, then cosine of this angle is AB/AC, the sine is BC/AC and tangent is BC/AB. In general sine of an angle = (opposite side)/(hypotenuse) cosine of an angle = (adjacent side)/(hypotenuse) tangent of an angle = (opposite side)/(adjacent side) |A | | | | | | |______________________C B
The union of the segments AB, BC, and AC of three nonlinear points A, B, and C.
If of triangle ABC and A'B'C' sides AB = A'B' and AC = A'C', and the included angle at A is larger than the included angle at A*, then BC > B'C'.Proof:A A'/|\ /|/ | \ / |/ | \ / |/ | \ B'/ |B | X \C |C'DWe construct AD such that AD = A'C' = AC and angle BAD = angle B'A'C'.Triangles ABD and A'B'C' are congruent. Therefore BD = B'C'.Let X be the point where the angle bisector of angle DAC meets BC.From the congruent triangles AXC and AXD (SAS) we have that XD = XC.Now, by the triangle inequality we have that BX + XD > BD, so BX + XC > BD, and consequently BC > BD = B'C'.
Area = 1/2*a*b*sin(C) where a = BC, b = AC and angle C = angle BCA 27 = 0.5*12*AC*sin(98) So AC = 27/[0.5*12*sin(98)] = 4.54 approx.
The definition of a triangle is the union of segments AB, BC, and AC (where A, B, and C are not all collinear).
It can be anything between zero and infinity, depending on the angle between AC and BC.
AC = 2*sqrt(3) = 3.4641AB = 2*sqrt(7) = 5.2915 angle ACB = 90 degrees.
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).