Question

Which classification best represents a triangle with side lengths 10 in., 12 in., and 15 in.? acute, because 102+122>152 acute, because 122+152>102 obtuse, because 102+122>152 obtuse, because 122+152>102

Answer 1

Answer:

(A) acute    

Step-by-step explanation:

We are given three side lengths that are:

m(A)=10 in

m(B)=12 in and

m(C)=15 in

Now, using the law of cosines, we get

$cosA=\frac{(12)^2+(15)^2-(10)^2}{2(12)(15)}$

$CosA=\frac{144+225-100}{360}$

$CosA=\frac{269}{360}$

$A=41.65^{\circ}$

Also, $CosB=\frac{(10)^2+(15)^2-(12)^2}{2(10)(15)}$

$CosB=\frac{181}{300}$

$B=52.89^{\circ}$

Therefore, ∠C=180-∠A-∠B

∠C=180-41.65-52.89

∠C=85.46°

which is an acute angle, therefore $10^{2}+12^{2}>15^{2}$.

Answer 2

We have three sides given such as:
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"