Trying to solve this problem

G

Guest

Guest
Hi there,

If any one can help me with this problem I would appreciate it:

"Let O be the circumcenter of Triangle ABC. Suppose AO intersects BD at D and that |AD| = 10, |BD| = 4, and |CD| = 5

Find sin A"


Here is what I have so far:
I draw the triangle inscribed in the circle. Note: A is on Top, B on bottom left, C on botom right, I also assume an acute triangle. Sides a = |BC|, b = |AC| and c = |AB|
Circumradius = R


Using extended law of sines: 2*R = a/Sin(A)

so Sin(A) = a / 2*R and since a = 9, we say Sin(A) = 9 /2R

So I am trying to figure out how to find the circumradius R.

Here are some things I have found from the picture:
2 * angle A = angle BOC by star trek lemma

triangle BOC is isoceles becuase sides |BO| and |CO| = R
Likewise triangles AOB and AOC are isoceles.

We cannot say AD is perpendicular to BC.

I feel like I am very close, but I am missing something very obvious, if anyone can help of give some hints I would be greatly appreciative.

thanks a lot
 
I found it

I got the solution,

Here it is for anyone interested,

Extend AD to E on the circle, then AE is the diameter.

Use power of the pont to find AE.

10*DE = 4*5 so DE = 2 and AE = 12

so circumradius is 6 and sinA = 9/12 or 3/4

If anyone attempted to solve this problem to help me out I greatly appreciate your help anyways.
 
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