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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))
What else can I help you with?
Who invented the cosine rule?
It was invented sometim but if you find out put it on here! Love Megan
What formula do you use to find the perimerter of a SSA triangle?
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
When doing trigonometry for example the cosine rule how do you know which side is a which side is b and which side is c if none of them are marked on and neither are the angles?
It's possible that either the angles or sides are labeled according to length or size.
Can trigonometric functions sine cosine and tangent be used in triangles other than right triangles?
YES!!! However, they are two functional equations. The Sine Rule is ;- SinA/a = SinB/b = SinC/c You select any two out of three above. The Sine Rule is used when two sides and two angles are being considered. The Cosine Rule is ; - a^(2) = b^(2) + c^(2) - 2bc CosA This rule is selected when three sides and one angle is being considered. NB Capital letter (A) is the angular value, and small/lower case letter (a) is the length of the opposite side. These equations/rules are used for scalene, Isosceles and equilateral triangles.
In the Sine and Cosine rule does it matter where the abc ABC go on the triangle for example can it go in any order or must it go in alphabetically order?
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").
CosA equals a2 plus c2 minus b2 by 2ac what is this formula?
This is known as the Cosine Rule.
How is the cosine rule derived?
the cosine rule is derived from the division of the adjacent side & hypotenuse
What is the formula for calculating the central angle given chord length and radius?
You can use the cosine rule to calculate the central angle.
How does one differentiate between the sine rule and the cosine rule?
The sine rule is a comparison of ratios: (sin A)/a = (sin B)/b = (sin C)/c. The cosine rule looks similar to the theorem of Pythagoras: c2 = a2 + b2 - 2ab cos C.
Who invented the cosine rule?
It was invented sometim but if you find out put it on here! Love Megan
What proportion do you use to find the length of the third side triangle XYZ?
The answer depends on the information that you have: it could be the sine rule or the cosine rule.
Is it is necessary to be a right angle triangle to use sine rule?
No. Sine rule (and cosine rule) apply to all triangles in Euclidean space (plane geometry). A simplification occurs when there is a right angle because the sine of the right angle is 1 and the cosine is 0. Thus you get Pythagoras theorem for right triangles.
How are all three versions of the law of cosines correct?
It follows from the cyclical symmetry of the cosine rule.
How does the exception not prove the rule?
The exception does not prove the rule because it contradicts the general principle or pattern that the rule represents. Instead of confirming the rule, the exception challenges its validity by showing that there are cases where the rule does not apply.
How do you find a measurement of an angle in a triangle when you only know the side length?
If you do not know only a side length you cannot. If you know all three side lengths then you can use the cosine rule. You can continue using the cosine rule for the other two angles but, once you have one angle, it is simpler to use the sine rule.
What is the radius of the circumcircle of a triangle with sides of 4cm by 5cm by 6cm?
Using the cosine rule the biggest angle is: 82.81924422 degrees Using radius formula for circumcircle of a triangle the radius is: 3.023715784 cm
What is the derivation of Heron's Formula?
The following proof is trigonometric, and basically uses the cosine rule. First we compute the cosine squared in terms of the sides, and then the sine squared which we use in the formula A=1/2bc·sinA to derive the area of the triangle in terms of its sides, and thus prove Heron's formula.We use the relationship x2−y2=(x+y)(x−y) [difference between two squares] [1.2]Finding the cosine squared in terms of the sidesFrom the cosine rule: We have:[1.3]Rearranging:[1.4]Because we want the sine, we first square the cosine:[1.5]Finding the SineTo use in: [1.6]Using Equation 1.5 in 1.6, we have:[1.7]Bringing all under the same denominator:[1.8]Using the difference between two squares (Equation 1.2)[1.9]Putting the above into a form where we can use the difference between two squares again we have:[1.10]Actually using the difference between two squares in both brackets, we find:[1.11]Substituting (a+b+c) for 2s, (b+c-a) for 2s-2a, etc:[1.12]Taking the square root:[1.13]Finding the AreaRecalling:[1.14]We have:[1.15]And simplified:
