Answer:
6.13
Explanation:
Using Sine Law we know that
![(a)/(SinA)=(b)/(SinB)=(c)/(SinC)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zgym0ghn1pkspjl7rnd3tzr66e4rog7q8g.png)
Using your figure let's assign sides and angles:
A=? B = 60° C = 70°
a = 5 b = ? c = x
If we put that into our formula:
![(5)/(Sin?)=(?)/(Sin60)=(x)/(Sin70)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2marh833i11v5iblvbgz543l7m81luljar.png)
Notice that we have too many unknowns. We need to complete at least one ratio to do this, so how do we do this?
Notice we have 2 angles given, so we solve for the third angle. The sum of all angles in any triangle is always 180°
∠A + ∠B + ∠C= 180°
∠A + 60° + 70° = 180°
∠A + 130° = 180°
∠A = 180° - 130°
∠A = 50°
Now we can use this to solve for x.
![(5)/(Sin50)=(x)/(Sin70)\\\\((5)(Sin70))/(Sin50) = x\\\\(4.6985)/(0.7660)=x\\\\6.1338 =x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ofdq0pjl2cvdjj6p0rv9x4weznau22puvl.png)
So the closest answer would be 6.13