Question

Solve a triangle with a =5, b =6, and c = 7. Round to the nearest tenth.

Answer 1

Answer:

<A ≈ 45 degrees

<B ≈ 57 degrees

<C ≈ 78 degrees

Step-by-step explanation:

Hi there!

1) Find <C with the law of cosines

Typically, we want to solve for the angle opposite the largest side first.

Law of cosines: $cosC=\frac{a^2+b^2-c^2}{2(a)(b)}$

Plug in given values

$cosC=\frac{5^2+6^2-7^2}{2(5)(6)}\\cosC=\frac{1}{5}\\C=cos^-^1(\frac{1}{5} )\\C=78$

Therefore, <C is approximately 78 degrees.

2) Find <B with the law of cosines

$cosB=\frac{a^2+c^2-b^2}{2(a)(c)}$

Plug in given values

$cosB=\frac{5^2+7^2-6^2}{2(5)(7)}\\cosB=\frac{19}{35}\\B=cos^-^1(\frac{19}{35})\\B=57$

Therefore, <B is approximately 57 degrees.

3) Find <A

The sum of the interior angles of a triangle is 180 degrees. To solve for <A, subtract <B and <C from 180:

180-57-78

= 45

Therefore, <A is 45 degrees.

I hope this helps!