+75 In △ABC, AB=2, D is a point on BC, such that AD bisects ∠BAC. If \(\frac{BD}{CD} = \frac{1}{3} \), ∠BAC=30∘, the value of BC^2 can be expressed as \(p + q\sqrt{r} \), where p,q, and r are integers, such that r is a positive integer free of square factors larger than 1. Find the value of p+q+r.
+128407 B
D
2
A C
Note that angles BAD and DAC = 15°
By the Law of Sines
sin BDA / 2 = sin BAD / BD
sin BDA / 2 = sin 15°/ BD
sin BDA = 2sin 15° / BD (1)
Since AD bisects BAC, then
AC / DC = AB/ BD → AC / 3BD = 2 /BD → AC = 3BD * 2 / BD = 6
And
sin CDA = sin BDA so
sin CDA / 6 = sin DAC / 3BD
sin BDA / 6 = sin 15° / 3BD → sin BDA = 6sin 15° / 3BD (2)
Equating (1) and (2)
2 sin 15° / BD = 6sin 15° / 3BD
3BD / BD = 6sin 15° / 2sin 15°
So
BC = 3BD + BD
BC^2 = (3B + BD)^2
BC^2 = ( 6sin 15° + 2sin 15°)^2 =
(sin 15° ( 6 + 2) )^2 =
sin^2 (15°) * 8^2 =
(sqrt 3 - 1)^2 / 8 * 64 =
(3 - 2sqrt 3 + 1) * 8 =
(4 - 2sqrt (3) ) * 8 =
32 - 16sqrt (3)
p = 32 q = -16 r = 3
p + q + r = 19
