+109 In triangle ABC, point D lies on side AC such that line segment BD bisects angle ABC. Angle A measures 30 degrees, angle C measures 45 degrees, and the length of side AC is 10 units. Determine the area of triangle ABD.
+9664 Note that \(\angle BAD = \angle CAD = \dfrac12 \cdot 30^\circ = 15^\circ\).
Considering the interior angle sum of \(\triangle ADC\) gives \(\angle ADC = 120^\circ\).
Using Sine Law in \(\triangle ADC \) gives
\(\dfrac{10}{\sin 120^\circ} = \dfrac{AD}{\sin 45^\circ}\\ AD = \dfrac{10 \sqrt 6}3\)
Considering the angles around point D gives \(\angle ADB = 60^\circ\). Considering the interior angle sum of \(\triangle ABD\) gives \(\angle ABD = 105^\circ\).
Using Sine Law in \(\triangle ABC\) gives
\(\dfrac{10}{\sin 105^\circ} = \dfrac{AB}{\sin 45^\circ}\\ AB = 10(\sqrt 3 - 1)\)
Then, the area of \(\triangle ABD\) is \(\dfrac12 (AB)(AD) \sin 15^\circ = \dfrac12 (10(\sqrt 3 - 1))\cdot \dfrac{10 \sqrt 6}3 \cdot \dfrac{\sqrt 6 - \sqrt 2}4 = \dfrac{50(2 \sqrt 3 - 3)}{3} \).
+128707 
Angle ABC =105
Let AD = x CD = 10 - x
BC / sin 30 = AC /sin 105
BC /sin30 = 10 /sin105
BC = 5/sin105
BA / sin 45 = AC / sin 105
BA / sin45 = 10 /sin105
BA = (10/sqrt 2) / sin 105 = 20 / (1 + sqrt 3)
Since BD is a bisector, then
BA / AD = BC / CD
(10/sqrt2)/ sin 105 / x = 5/sin105 / (10-x)
(10/sqrt 2) (10-x) / sin 105 = 5x / sin105
(10/sqrt 2) (10 -x) = 5x
100/sqrt 2 - (10/sqrt 2)x = 5x
100/sqrt 2 = (5 + 10/sqrt 2) x
x = (100/sqrt 2) / ( 5 + 10/sqrt 2) = (100) / (5sqrt 2 + 10) = 20 / (2 + sqrt 2) = AD
[ABD ] = (1/2)BA * AD * sin 30 = (1/4) [20 / (1 + sqrt 3)] * [ 20 /(2 + sqrt 2) ] ≈ 10.72
