In right triangle $ABC$, we have $\angle BAC = 90^\circ$ and $D$ is the midpoint of $\overline{AC}$. If $AB = 7$ and $BC = 7 \sqrt{3}$, then what is $\tan \angle BDC$?
+128707 B
7 7sqrt (3) = sqrt (147)
A D C
AC = sqrt [ 147 - 49 ] = sqrt [98 ] = 7sqrt (2)
DC = 3.5sqrt (2) = AD = sqrt [24.5]
BD = sqrt [ 7^2 + 24.5] = sqrt [ 73.5]
Using Law of Cosines
BC^2 = BD^2 + DC^2 - 2(BD* DC) cos (BDC)
147 = 73.5 + 24.5 - 2 [ sqrt [ 73.5] * sqrt [ 24.5] ] cos (BDC)
49 / [- 2 sqrt (7203 / 4 ) ] = cos (BDC)
-49 / sqrt(7203) = cos (BDC)
sin(BDC) = sqrt [ 1 - (49)^2 / 7203 ] = sqrt (6) / 3
tan BDC = [sqrt (6) / 3 ] / [ -49 / sqrt (7203) ] = [sqrt(6)*sqrt (7203) / [ -49 * 3] =
(sqrt 2) (sqrt (3) * 49 * sqrt (3) ) / [ -147] =
-sqrt (2) * 147 / 147 =
-sqrt (2)
