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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

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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.




In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft-example-1
User Trewaters
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4 votes

Answer:

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

In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft-example-1
User Shayki Abramczyk
by
7.5k points

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