Answer:
The length of the angle bisector of angle ∠A is 6.01.
Step-by-step explanation:
It is given that length of leg AC = 5 ft and the hypotenuse AB = 13 ft.
Using pythagoras theorem
[tex](AB)^2=(BC)^2+(AC)^2[/tex]
[tex](13)^2=(BC)^2+(5)^2[/tex]
[tex]169=(BC)^2+25[/tex]
[tex]BC=12[/tex]
[tex]\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}[/tex]
[tex]\sin A=\frac{BC}{AB}[/tex]
[tex]A=\sin ^{-1}\frac{12}{13}[/tex]
[tex]A=67.38[/tex]
Bisector divides the angle in two equal parts, therefore,
[tex]A'=\frac{67.38}{2} =33.69[/tex]
In triangle ACD.
[tex]\cos A'=\frac{\text{Base}}{\text{Hypotenuse}}[/tex]
[tex]\cos A'=\frac{AC}{AD}[/tex]
[tex]\cos (33.69^{\circ})=\frac{5}{AD}[/tex]
[tex]0.832=\frac{5}{AD}[/tex]
[tex]AD=\frac{5}{0.832} =6.009\approx 6.01[/tex]
Therefore the length of the angle bisector of angle ∠A is 6.01.
