Answer :
The required length of AC in triangle ABC is b = 119.57 units.
How to find the value of AC?
In triangle ABC, we have:
[tex]$\angle B = 85^\circ$[/tex]
[tex]$\angle C = 53^\circ$[/tex]
[tex]$BC = 85 units[/tex]
AB = c
AC = b (we need to find the value of b)
We know that the sum of the interior angles of a triangle is [tex]$180^\circ$[/tex]. Therefore:
[tex]$\angle A = 180^\circ - \angle B - \angle C$[/tex]
[tex]$\angle A = 180^\circ - 85^\circ - 53^\circ$[/tex]
[tex]$\angle A = 42^\circ$[/tex]
Using the law of sines, we have:
[tex]$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$[/tex]
Substituting the given values, we get:
[tex]$\frac{c}{\sin 85^\circ} = \frac{b}{\sin 53^\circ}$[/tex]
Solving for $c$, we get:
[tex]$c = b \cdot \frac{\sin 85^\circ}{\sin 53^\circ}$[/tex]
We also know that:
[tex]$\frac{a}{\sin A} = \frac{85}{\sin 42^\circ}$[/tex]
Solving for a, we get:
[tex]$a = 85 \cdot \sin 42^\circ$[/tex]
Now we can use the law of cosines to find b:
[tex]$b^2 = a^2 + c^2 - 2ac \cdot \cos B$[/tex]
Substituting the known values, we get:
[tex]$b^2 = (85 \cdot \sin 42^\circ)^2 + \left[b \cdot \frac{\sin 85^\circ}{\sin 53^\circ}\right]^2 - 2 \cdot 85 \cdot b \cdot \sin 42^\circ \cdot \sin 85^\circ/(\sin 53^\circ)$[/tex]
Simplifying and solving for b, we get:
b = 119.57 units
Therefore, the length of AC in triangle ABC is b = 119.57 units.
To know more about Triangle visit:
brainly.com/question/2773823
#SPJ1
