Question

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

Answer 1

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.

Answer 2

AD= 6 ft

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

`(AB)^2=(BC)^2+(AC)^2`

⇒
`(13)^2=(BC)^2+(5)^2`

⇒
`169=(BC)^2+25`

⇒
`169-25=(BC)^2`

⇒
`144=(BC)^2`

⇒ BC=12

Now, we find the ∠A

`sinA= (Perpendicular)/(Hypotenuse)`

`sinA= (BC)/(AB)`

`sinA= (12)/(13)`

`A= sin^(-1)(12)/(13)`

∠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

`CosA'= (Base)/(Hypotenuse)`

`CosA'= (AC)/(AD)`

`Cos(33.69)= (5)/(AD)`

`0.832=(5)/(AD)`

`AD=(5)/(0.832)`

`AD=6.009`

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