# The diameter of a circle.

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

### The diameter of a circle.

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?

1)explanation

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.
Guest

### Re: The diameter of a circle.

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

HallsofIvy

Posts: 143
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 57

Return to Trigonometry - sin, cos, tan, cot, arcsin, arccos, arctan, arccot

### Who is online

Users browsing this forum: No registered users and 3 guests