in ABC, AD is the angle bisector of /_BAC.
What is BD?
Enter your answer as a decimal,
? in.

Answer:
1.8
Step-by-step explanation:
ADC is a right angled triangle. So is ADB.
For right angled triangle ADC,
[tex] {ad}^{2} + {dc}^{2} = {ac}^{2} \\ {x}^{2} + {3}^{2} = {4}^{2} \\ {x}^{2} + 9 = 16 \\ {x}^{2} = 16 - 9 = 7 \\ x = \sqrt{7} [/tex]
For right angled triangle ADB,
[tex] {ad}^{2} + {db}^{2} = {ab}^{2} \\ 7 + {y}^{2} = {3.2}^{2} \\ 7 + {y}^{2} = 10.24 \\ {y}^{2} = 10.24 - 7 \\ {y}^{2} = 3.24 \\ y = \sqrt{3.24} \\ y = 1.8[/tex]