Respuesta :

The distances from D to the both sides(AB and AC) of the triangle are 4 ft.

Explanation

In the diagram below, ABC is a triangle in which ∠CAB = 60°

As,  AD is the angle bisector of ∠CAB, that means ∠CAD = ∠DAB = 30°

The distance from D to side AC is  DE and distance from D to side AB is  DF. That means both ∠AED and ∠AFD are right angle.

Given that, the length of AD = 8 ft.

So, in right angle triangle AED.....

[tex]sin(30)= \frac{DE}{AD}\\ \\ sin(30)= \frac{DE}{8} \\ \\ \frac{1}{2}= \frac{DE}{8}\\ \\ 2*DE=8\\ \\ DE=4[/tex]

Also, in right angle triangle AFD....

[tex]sin(30)= \frac{DF}{AD}\\ \\ sin(30)= \frac{DF}{8} \\ \\ \frac{1}{2}= \frac{DF}{8}\\ \\ 2*DF=8\\ \\ DF=4[/tex]

So, the distances from D to the both sides(AB and AC) of the triangle are 4 ft.

Ver imagen Sicista

The distances from [tex]D[/tex] to the both sides([tex]AB[/tex] and [tex]AC[/tex]) of the triangle are [tex]4\;\rm{ft[/tex].

Step-by-step explanation:

Given: In △ABC, [tex]\angle CAB=60^\circ[/tex] and [tex]AD[/tex] is angle bisector with [tex]D\in{BC}\;\rm{and }\; AD=8\;\rm{ft[/tex].

From the figure,

[tex]\angle BAD=\angle DAC=30^\circ[/tex]

In [tex]\Delta AED, \;\angle E=90^\circ[/tex]

[tex]sin30^\circ=\frac{DE}{AD}[/tex]

        [tex]\\\frac{1}{2}=\frac{DE}{8}\\[/tex]

   

In [tex]\Delta AFD, \;\angle F=90^\circ[/tex]

[tex]sin30^\circ=\frac{DF}{AD}[/tex]

        [tex]\\\frac{1}{2}=\frac{DF}{8}\\[/tex]

    [tex]DF=4\;\rm{ft[/tex]

Hence, the distances from to the both sides ( and ) of the triangle are .

Learn more from trigonometric ratios here:

https://brainly.com/question/23641403?referrer=searchResults

Ver imagen divyamadaan8054