Question

PLEASE HELP.

Answer 1

We know that
the law of cosines establish 
b²=a²+c²-2*a*c*cos B
in this problem
a=4 units
c=5 units
B=60°
so
b²=a²+c²-2*a*c*cos B
b²=4²+5²-2*4*5*cos 60°
b²=16+25-40*(1/2)
b²=16+25-20
b=√[16+25-20]

the answer is
b=√[16+25-20]

Answer 2

Answer:

(B) $b=\sqrt{25+16-20}$

Step-by-step explanation:

It is given that we have to consider triangle ABC in which ∠B=60° and the side lengths are a=4 and c=5.

Therefore, in order  to find the value of the side length b, we will use the Law of Cosines that is:

$b^2=c^2+a^2-2(c)(a)cosB$

substituting the given values, we get

$b^2=(5)^2+(4)^2-2(5)(4)cos60^{\circ}$

$b^2=25+16-40(\frac{1}{2})$

$b^2=25+16-20$

$b=\sqrt{25+16-20}$

which is the required side length of b.

Thus, option B is correct.