The diameter of a circle.

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

The diameter of a circle.

Postby Guest » Wed Nov 28, 2018 8:05 pm

The inverse of one angle and the side opposite to the angle, which is equal to one, in all triangles gives a diameter of the circumscribed circle.

In a right triangle the inverse of the angle is always 1 which is the circle's diameter, and its' opposite's side is 1, that is hypotenuse of 1. $\sin 90 =1$ is the angle $\theta$.

My question is, what is the right explanation concerning the inverse of angles with sides $a,b,c$ for all $\triangle ABC$ when the base or the hypotenuse equals to 1?


I have three sides $5, 5, 4$ and $1.25, 1.25, 1$ with all the same angles except the base is 1, and all I did is divide sides $5, 5, 4$ by 4 to obtain the the simplyfied version of the triangle sides. $1.25, 1.25, 1$ . From the $\triangle ABC $ and 1 the side $c$ opposite to the $\angle C =\sqrt (1-0.68^2)$. Its' inverse is the diameter of circumscribed circle in term of $\sin$ but not in terms of Cosine.

The law of Sine states that :

$$\frac {\sin A}{a}=\frac {\sin B}{b}=\frac {\sin C}{c}=\frac {1}{d}$$ where d is the diameter.

$$\frac {a}{\sin A}=\frac {b}{\sin B}=\frac {c}{\sin C}={d}$$ where d is the diameter.

Re: The diameter of a circle.

Postby HallsofIvy » Thu Dec 19, 2019 5:28 pm

I think you mean "reciprocal" rather than "inverse". Also the sine law says that [tex]\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}[/tex] but does NOT include [tex]\frac{1}{d}[/tex]. For example, an equilateral triangle inscribed in a unit circle has [tex]sin(A)= sin(B)= sin(C)= sin(\pi/3)= \sqrt{3}/2[/tex] and [tex]a= b= c= \sqrt{3}. [tex]\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}= \frac{\frac{\sqrt{3}}{2}}{\sqrt{3}}= \frac{1}{2}[/tex], NOT 1.

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