Thread: Isosceles' triangle circumcircle radius

Hi,

This should help you get started :

First, you know (or should know) that the main height AH in the isosceles triangle ABC cuts the segment BC in half (because the triangle is isosceles and the Ah is the main height). So HC=BH=8. You can see that AH=1/2 HB.

Now, finding angles :

trigonometry : Because the triangle BHA is right-angled we can use the main sin, cos, tan (if not, we would have used the sin law* or the cos law).

So tan(B) = AH/HB = 1/2 so B=tan^-1 (1/2) = 26.57 degrees so C=26.57 degrees and A=180-2*26.57=126.87 degrees

Have fun !!

* the sin law s the part of what you stated in your problem which is sin(A)/a=sin(B)/b=sin(C)/c where a is the side facing the angle A and b the side facing angle B... This law and the cosin law work in every triangle.