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In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: The length of angle bisecor of angle A

Respuesta :

If AC= 5ft, AB=3ft, then we can know that BC²=AB²-AC²， BC=12ft. We draw an angle bisector AD of angle A, and draw a vertical line of AB at point D, that's segment DE. Because of ∠CAD=∠EAD, △CAD≌△EAD. That means CD=DE, AC=AE=5ft, so BE=AB-AE=13-5=8ft. If CD=xft=DE, BD=(12-x)ft. According to the Pythagorean theorem, DE²+BE²=BD²，we can create a equation.                 x²+8²=(12-x)²        x=10/3 ft                                                  Then, we use Pythagorean theorem again , AC²+CD²=AD²，and we can get the length of angle bisector of angle A.




Ver imagen rwang

Answer:

AD= 6 ft

Step-by-step explanation:

We have given that : In a right triangle ΔABC, the length of leg AC = 5 ft               and the hypotenuse AB = 13 ft

To find : The length of angle bisector of angle A

Solution : Since, ΔABC is a right angle triangle

applying Pythagoras theorem, we can find the length of side BC

    [tex](AB)^2=(BC)^2+(AC)^2[/tex]

⇒ [tex](13)^2=(BC)^2+(5)^2[/tex]

⇒ [tex]169=(BC)^2+25[/tex]

⇒ [tex]169-25=(BC)^2[/tex]

⇒ [tex]144=(BC)^2[/tex]

 ⇒ BC=12

Now, we find the ∠A

[tex]sinA= \frac{Perpendicular}{Hypotenuse}[/tex]

[tex]sinA= \frac{BC}{AB}[/tex]

[tex]sinA= \frac{12}{13}[/tex]

[tex]A= sin^{-1}\frac{12}{13}[/tex]

∠A=67.38°

Now, we have given that A is the angle bisector on BC at pt. D

which means it divide angle into two equal parts

Therefore, ∠A'= ∠A /2= 67.38/2= 33.69°

Now, In ΔCAD

[tex]CosA'= \frac{Base}{Hypotenuse}[/tex]

[tex]CosA'= \frac{AC}{AD}[/tex]

[tex]Cos(33.69)= \frac{5}{AD}[/tex]

[tex]0.832=\frac{5}{AD}[/tex]

[tex]AD=\frac{5}{0.832}[/tex]

[tex]AD=6.009[/tex]

Therefore, the length of the angle bisector of ∠A = AD≈ 6 ft

Ver imagen DodieZollner

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