Question

Suppose a triangle has sides a, b, and c, and let (/) be the angle opposite the side of length a. If cos(/) > 0 what must be true?
A. b^2 + c^2 < a^2
B. a^2 + b^2 = c^2
C. b^2 + c^2 > a^2
D. a^2 + b^2 > c^2

Answer 1

   Using the cosine rule

 a² =  b² +  c² -2 bc cos (/)

 Simplifying the equation in terms of cos (/),

 cos (/)= (b² + c² - a²)/(2bc) 

 If cos(/) > 0

 (b² + c² - a²)/(2bc) > 0 -----------------(b² + c² - a²) > 0 

the resulting inequality should be

(b² + c² ) > a² 

The answer is the letter C. b^2 + c^2 > a^2

Answer 2

cos(/)>0, that implies that a^2+b^2!=c^2. If that were the case, cos(/) would be 0. cos(/)>0 implies that the adjacent is greater than the hypotenuse. So, the opposite plus the adjacent is greater than the hypotenuses. So, then, that implies b^2+a^2>c^2. So, A.