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

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boffeemadrid boffeemadrid · Answer 1 · 2019-02-27T09:41:42+01:00

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

[tex]cosA=\frac{(12)^2+(15)^2-(10)^2}{2(12)(15)}[/tex]

[tex]CosA=\frac{144+225-100}{360}[/tex]

[tex]CosA=\frac{269}{360}[/tex]

[tex]A=41.65^{\circ}[/tex]

Also, [tex]CosB=\frac{(10)^2+(15)^2-(12)^2}{2(10)(15)}[/tex]

[tex]CosB=\frac{181}{300}[/tex]

[tex]B=52.89^{\circ}[/tex]

Therefore, ∠C=180-∠A-∠B

∠C=180-41.65-52.89

∠C=85.46°

which is an acute angle, therefore [tex]10^{2}+12^{2}>15^{2}[/tex].

MrRoyal MrRoyal · Answer 2 · 2022-01-31T17:38:37+01:00

The triangle is an acute triangle because 10^2 + 12^2 > 15^2

The side lengths of the triangles are:

10, 12 and 15

The smallest side lengths are: 10 and 12

The sum of the squares of these side lengths is

[tex]10^2 + 12^2[/tex]

For the triangle to be an acute triangle, then the following must be true

[tex]10^2 + 12^2 > 15^2[/tex]

Evaluate the exponents

[tex]244 > 225[/tex]

The above inequality is true.

Hence, the triangle is an acute triangle because 10^2 + 12^2 > 15^2