In $\triangle ABC,AD$ and $BE$ are the medians drawn through the angular points $A$ and $B$ respectively. $\angle{DAB}=2\angle{ABE}=36^{\circ}$ and $AD=6$ units, then circumradius of the triangle is equal to
- $(3-\sqrt{5}) \text{cosec} C$
- $(3+\sqrt{5}) \text{cosec}C$
- $2(3-\sqrt{5}) \text{cosec}C$
- $2(3+\sqrt{5}) \text{cosec}C$
Using the properties of medians and the given angles, the triangle geometry can be solved using the sine rule and median length formulas. The circumradius R is calculated as (3+sqrt(5)) * cosec(C).
Since AD and BE are medians, D and E are midpoints of BC and AC respectively. In triangle ABD, we know angle DAB is 36 degrees and side BD is half of BC, while in triangle ABE, angle ABE is 18 degrees and side AE is half of AC. Using the sine rule in triangle ABD, we get BD divided by sin(angle DAB) equals AD, so a/2 divided by sin(36 degrees) equals 6. This gives a = 12 sin(36 degrees). Using the sine rule again in triangle ABE, we get AE divided by sin(angle ABE) equals BE, so b/2 divided by sin(18 degrees) equals BE. In triangle ABE, angle AEB is 180 degrees minus (18 degrees plus 54 degrees) equals 108 degrees, so BE divided by sin(54 degrees) equals b divided by sin(108 degrees). Since sin(108 degrees) equals sin(72 degrees), BE becomes b sin(18 degrees) divided by sin(72 degrees), which is b divided by (2 cos(18 degrees)). Equating the two expressions for BE yields 6 divided by cos(18 degrees) equals b divided by (2 cos(18 degrees)), meaning b = 12. Using the extended law of sines in triangle ABC, the circumradius R equals c divided by (2 sin C). By the cosine rule, c squared equals 144 plus 144 sin squared(36 degrees) minus 144 sin(36 degrees) cos(36 degrees). Because cos(36 degrees) minus sin(18 degrees) equals (sqrt(5) minus 1)/2 minus (sqrt(5) minus 1)/4, we have 2(cos(36 degrees) minus sin(18 degrees)) equals (sqrt(5) minus 1), which makes the side length c equal to 6(sqrt(5) minus 1). Therefore, R = (3 + sqrt(5)) cosec C.