Question

For the adjoint of the
product of two operators
A and B​

Answer 1

Explanation:

What is the circumradius of a triangle when its vertices are (0,0), (3,0) and (1,4/3)?

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The vertices of the triangle are A(0,0),B(3,0) and C(1,43).

⇒ The lengths of the sides of the triangle are

BC=a=√4+169−−−−−√=52√3,

AC=b=1+169−−−−−√=53, and

AB=c=3.

The length of the altitude from vertex C to side AB is 43.

⇒ The area of △ABC is 12(base×height)=12(3×43)=2.

The formula for the circumradius of a triangle is R=abc4A, where a,b and c are t

R2=a2b2c216Δ2

We have squared sides

32=9

12+(4/3)2=25/9=52/32

(3−1)2+(4/3)2=52/9=22(13)/32

We can get the area any number of ways; base is 3 along x axis, height 4/3, so area

Δ=12(3)(4/3)=2

R2=(32)(52/32)(22(13)/32)16(22)

R2=52(13)4232

R=51213−−√