Triangle 75

Triangle ABC has angle C bisected and intersected AB at D. Angle A measures 20 degrees, and angle B measures 40 degrees. The question is to determine AB-AC if length AD=1.

Correct answer:

x =  0.3949

Step-by-step explanation:

$$\begin{gathered} AD=1 \\ A=20{}^{\circ } \\ B=40{}^{\circ } \\ C=180-A-B=180-20-40=120{}^{\circ } \\ \angle ACD=C/2=120/2=60{}^{\circ } \\ \angle DCB=C/2=120/2=60{}^{\circ } \\ \angle ADC=180-A-\angle ACD=180-20-60=100{}^{\circ } \\ \angle CDB=180-B-\angle DCB=180-40-60=80{}^{\circ } \\ \frac{\sin ADC}{\sin ACD}=AC:AD \\ AC=AD\cdot \frac{\sin ADC}{\sin ACD}=AD\cdot \frac{\sin 100°}{\sin 60°}=1\cdot \frac{\sin 100°}{\sin 60°}=1\cdot \frac{0.984808}{0.866025}=1.13716 \\ DC=AD\cdot \frac{\sin A}{\sin ACD}=AD\cdot \frac{\sin 20°}{\sin 60°}=1\cdot \frac{\sin 20°}{\sin 60°}=1\cdot \frac{0.34202}{0.866025}=0.39493 \\ DB=DC\cdot \frac{\sin DCB}{\sin B}=DC\cdot \frac{\sin 60°}{\sin 40°}=0.3949\cdot \frac{\sin 60°}{\sin 40°}=0.3949\cdot \frac{0.866025}{0.642788}=0.53209 \\ AB=AD+DB=1+0.5321\doteq 1.5321 \\ x=AB-AC=1.5321-1.1372=0.3949 \end{gathered}$$

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