$$ABC$$ is a right angled triangle. $$\angle BAC = 90$$° and $$\angle ACB = 60$$°. What is the ratio of the circum radius of the triangle to the side $$AB?$$
In a right angled triangle circum radius is half of the hypotenuse
AC/2 =R
AC=2R
Also Sin 60=AB/AC
$$\sqrt{3}/2$$=AB/AC
$$\sqrt{3}/2$$=AB/2R
R/AB=1:$$\sqrt{3}$$
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