Equlateral Triangle: Prove $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$

• MHB
• lfdahl
In summary, the conversation discusses two different ways to show that a triangle is equilateral if and only if the product of two sides and the cosine of the opposite angle is equal for all three angles. Method 1 involves using the given values of the cosine, while method 2 is not specified.
lfdahl
Gold Member
MHB
Given a triangle with angles $\alpha, \beta, \gamma$ opposite respectively to the sides $a,b,c$. Show in at least two different ways, that the triangle is equilateral if and only if $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$.

lfdahl said:
Given a triangle with angles $\alpha, \beta, \gamma$ opposite respectively to the sides $a,b,c$. Show in at least two different ways, that the triangle is equilateral if and only if $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$.

Method 1)

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
hence $\frac{1}{c}\cos \gamma =\frac{1}{b} \cos \beta = \frac{1}{a} \cos \alpha$
we have from sin rule $\frac{1}{c}\sin \gamma =\frac{1}{b} \sin \beta = \frac{1}{a} \sin \alpha$
square and add to get $\frac{1}{c^2} =\frac{1}{b^2} = \frac{1}{a^2}$ or $a=b=c$ hence equilateral

for the other part if $a=b=c$ then
$\alpha = \beta = \gamma$
hence $\cos \alpha = \cos \beta = \cos \gamma$ and by multiplying we get the result

method 2

using the values of cos we get
$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
$<==> a^2+b^2-c^2 = a^2+c^2-b^2= b^2 + c^2 - a^2$
from $a^2+b^2-c^2 = a^2+c^2-b^2$
$<==> b^2-c^2<==>b=c$ simlilarly a =c $<==>$ it is equilateral

Method 1)

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
hence $\frac{1}{c}\cos \gamma =\frac{1}{b} \cos \beta = \frac{1}{a} \cos \alpha$
we have from sin rule $\frac{1}{c}\sin \gamma =\frac{1}{b} \sin \beta = \frac{1}{a} \sin \alpha$
square and add to get $\frac{1}{c^2} =\frac{1}{b^2} = \frac{1}{a^2}$ or $a=b=c$ hence equilateral

for the other part if $a=b=c$ then
$\alpha = \beta = \gamma$
hence $\cos \alpha = \cos \beta = \cos \gamma$ and by multiplying we get the result

method 2

using the values of cos we get
$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
$<==> a^2+b^2-c^2 = a^2+c^2-b^2= b^2 + c^2 - a^2$
from $a^2+b^2-c^2 = a^2+c^2-b^2$
$<==> b^2-c^2<==>b=c$ simlilarly a =c $<==>$ it is equilateral

1. What is an equilateral triangle?

An equilateral triangle is a type of triangle where all three sides are equal in length and all three angles are equal at 60 degrees.

2. How can you prove that $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$ in an equilateral triangle?

This equation can be proven using the Law of Cosines, which states that for any triangle with sides a, b, and c and angles $\alpha$, $\beta$, and $\gamma$, the following equation holds true: $c^2 = a^2 + b^2 - 2ab\cos \gamma$. In an equilateral triangle, all three angles are equal at 60 degrees, so the equation becomes $c^2 = a^2 + b^2 - 2ab\cos 60$. Since in an equilateral triangle, all sides are equal, we can substitute a, b, and c with just one variable, x, and we get $x^2 = x^2 + x^2 - 2x^2 \cos 60$. Simplifying, we get $1 = 1 - \cos 60$, which is true. Therefore, we have proven that $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$ in an equilateral triangle.

3. What is the significance of this equation in an equilateral triangle?

This equation helps to show the relationship between the sides and angles in an equilateral triangle. It also shows that in an equilateral triangle, the cosine of one angle is equal to the cosine of the other two angles, which is unique to equilateral triangles.

4. Can this equation be used to find the length of the sides or measure of the angles in an equilateral triangle?

Yes, this equation can be used to find the length of the sides or measure of the angles in an equilateral triangle if at least two of the values are known. For example, if you know the length of two sides, you can use the equation to find the measure of the third side. Or if you know the measure of one angle, you can use the equation to find the measure of the other two angles.

5. Does this equation hold true for all triangles or only equilateral triangles?

This equation only holds true for equilateral triangles. In other types of triangles, the angles and sides are not equal, so the equation does not apply.

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