Question

Which classification best represents a triangle with side lengths 10 in., 12 in., and 15 in.?

Answer 1

mA = 10 in

mB =12 in

mC =15 in

Solving for the angles using the law of Cosines:

cosA = (12² + 15²- 10²)/2*12*15

A = 41.65°

cosB = (10²+15²-12²)/2*10*15

B = 52.89°

C = 180° - 41.65° - 52.89°

C = 85.46° 

The Answer is

10² + 12² > 15² acute

Answer 2

side length of ∆ABC , a = 10 , b= 12 and c = 15
use cosine formula to get angle between two sides,
$cosA=\frac{b^2+c^2-a^2}{2bc}$
cosA = (12² + 15² - 10²)/2.12.15
= (144 + 225 - 100)/360
= (144 + 125)/360
= 169/360

similarly, $cosB=\frac{c^2+a^2-b^2}{2ca}$
cosB = (15² + 10² - 12²)/2.10.15
= (225 + 100 - 144)/300
= (225 - 44)/300
= 181/300

$cosC=\frac{a^2+b^2-c^2}{2ab}$
cosB = (10² + 12² - 15²)/2.10.12
= (100 + 144 - 225)/240
=19/240

here ,we see cosA , cosB , cosC all are positive means value of A, B, C less then 90° but greater than 0°. hence definitely this triangle is acute angled triangle.