December 1st, 2018, 04:33 AM  #1 
Newbie Joined: Dec 2018 From: Canada Posts: 1 Thanks: 0 
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 simplified version of the triangle sides. $1.25, 1.25, 1$. From the $\triangle ABC $ and 1, the side $c$ opposite to the $\angle C =\sqrt {10.68^2}$. Its inverse is the diameter of circumscribed circle in terms of $\sin$, but not in terms of cosine. The law of sines 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. Last edited by skipjack; December 1st, 2018 at 08:52 AM. 

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