+0

# Geometry problem

0
16
3
+296

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.

May 1, 2024

#1
+9665
+1

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 $$\dfrac12 (AB)(AD) \sin 15^\circ = \dfrac12 (10(\sqrt 3 - 1))\cdot \dfrac{10 \sqrt 6}3 \cdot \dfrac{\sqrt 6 - \sqrt 2}4 = \dfrac{50(2 \sqrt 3 - 3)}{3}$$.

May 1, 2024
#3
+9665
0

This is wrong, but I can't edit it, see CPhill's answer below

MaxWong  May 2, 2024
#2
+129829
+1

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

May 1, 2024