# Which is the type of the triangle where: (3^1/2)*a-(3^1/2)*b=c a+b=(3^1/2)*c a,b,c=the length of each side of the triangle

*print*Print*list*Cite

### 2 Answers

From the given conditions we get (a-b)=c/sqrt3 and a+b=c(sqrt3);

Sum and difference of the squares ,divided by 2 gives:

**a^2+b^2=(5/3)c^2** and **2ab=(4/3)c^2**and, **(a^2-b^2**=(a+b)(a-b)=**c^2**

By trigonometry,

a^2=b^2+c^2-2bccosA, where A is the angle between a and b.

cosA= {(b^+c^2)-a^2}/(2bc)={c^2-(a^2-b^2)}/(2ab), Substite a^2-b^2 with its value found above

cosA={(c^2-(a^2+b^2)}/2bc=(c^2-c^2)/(2bc)=0

Therefore, **angle A is 90 degree.**

CosC= {a^2+b^2-c^2}/(2ab)={(5/3)c^2-c^2}/(4/3)c^2=(2/3)(3/4)=1/2.

Therefore angle C=cos inverse(1/2)=**60 degree.**

The remaining angle B = 180-(A+C)=180(90+60)=**30** degree.

Sides:

angle C= 30=>** b**= c(cot60)= **(1/sqrt3)c**

**a**=c(sec60)=**(2/sqrt3)c**.

The right angle identified is as follows with its parameters:

A=90, B=30, C=30 degree

a=(2/sqrt3)c , b= (1/sqrt3)c, c= 1c

We'll try to determine the length of b and the length of c, considering the length of a side.

We'll multiply the second relation with the value (3^1/2) and after that we'll add the equivalent obtained relation to the first one.

(3^1/2)*a + (3^1/2)*b + (3^1/2)*a - (3^1/2)*b = 3*c + c

We'll group the same terms:

2*(3^1/2)*a = 4*c

(3^1/2)*a = 2 *c

**c= [(3^1/2)*a]/2**

With the c value written in function of "a" value, we'll go in the second relation and substitute it:

a + b = [(3^1/2)*(3^1/2)*a]/2

a + b = 3*a/2

We'll have the same denominator on the left side of the equality:

2*a + 2*b = 3*a

2*b = 3*a - 2*a

2*b = a

**b = a/2**

If the triangle is a right one, then, using the Pythagorean theorem, we'll have the following relation between the sides of triangle:

a^2 = b^2 + c^2

Now, we have to plug in the values of "b" and "c", in the relation above:

a^2 = a^2/4 + 3*a^2/4

a^2 = 4*a^2/4

**a^2 = a^2**

We've shown that the equality is true, so the triangle is right, where "a" is hypotenuse and "b","c" are cathetus.