We stat with the law of cosines, which we can assume to be true:
* c2 = a2 + b2 - 2ab*cos(C) Then rearrange it:
* cos(C) = a2 + b2 - c2/2ab Use the identity sin(x)=SQRT(1-cos(x))
* sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2) Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out.
* 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab ab's cancel, and the term inside the square root is the difference of two squares.
* A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2))) when the two groups are simplified, the can be factored in binomial squares.
* A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2) Once again, we have differences of squares.
* A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c)) Simplify.
* A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c)) Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s.
* A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s) The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula.
* A = SQRT(s(s - a)(s - b)(s - c))
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An SSA triangle is ambiguous.Suppose the triangle is ABC and, with conventional labelling, you know a, b and angle A.Then by the cosine rule, a2 = b2 + c2 - 2bc*Cos(A)This equation will give rise to a quadratic equation in cwhich has 2 solutions. The perimeter is then a + b + c1 or a + b + c2
It's possible that either the angles or sides are labeled according to length or size.
Yes, trigonometric functions such as sine, cosine, and tangent can be applied to triangles other than right triangles through the use of the Law of Sines and the Law of Cosines. These laws relate the ratios of the sides of any triangle to the sines and cosines of their angles, allowing for the calculation of unknown sides and angles in non-right triangles. Thus, trigonometric functions are versatile tools applicable to various types of triangles.
The order of them does not matter at all, as long as the sides are consistently opposite the angles with the corresponding letter (e.g. side "A" is always opposite angle "a").