Given triangle ABC with B = 37°, a = 25 feet, and b = 16 feet, if the law of cosines is used to solve the triangle...

Question

Given triangle ABC with B = 37°, a = 25 feet, and b = 16 feet, if the law of cosines is used to solve the triangle, what quadratic equation must first be solved?

A) $c^2 + 39.93c − 881 = 0$

B) $c^2 + 25.56c + 881 = 0$

C) $c^2 − 39.93c + 369 = 0$

D) $c^2 − 25.56c − 369 = 0$

When I try solving this I get:

$$b^2=a^2+c^2-2ac(cosB)$$ $$(16)^2=(25)^2+c^2-2(25)c(cos(37))$$ $$0=(-16)^2+(25)^2+c^2-2(25)c(cos(37))$$ $$0=-39.93c+c^2+881$$ $$c^2-39.93c+881 $$

Answer 1

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

$$(16)^2=(25)^2+c^2-2(25)c(cos(37))$$

$$0=(25)^2+c^2-2(25)c(cos(37))-(16)^2$$