In triangle ABC, a=4,b=6, c=8, how do you find the cosine of each of the angles?

Jan 10, 2017

Use 3 forms of Law of Cosines :
${a}^{2} = {b}^{2} + {c}^{2} - 2 \left(b\right) \left(c\right) \cos \left(A\right)$
${b}^{2} = {a}^{2} + {c}^{2} - 2 \left(a\right) \left(c\right) \cos \left(B\right)$
${c}^{2} = {a}^{2} + {b}^{2} - 2 \left(a\right) \left(b\right) \cos \left(C\right)$
Solve for cosines

Explanation:

The first form of the Law of Cosines :

${a}^{2} = {b}^{2} + {c}^{2} - 2 \left(b\right) \left(c\right) \cos \left(A\right)$

Solve for cos(A):

$\cos \left(A\right) = \frac{{a}^{2} - {b}^{2} - {c}^{2}}{- 2 b c}$

Substitute, 4 for a, 6 for b, and 8 for c:

$\cos \left(A\right) = \frac{{4}^{2} - {6}^{2} - {8}^{2}}{- 2 \left(6\right) \left(8\right)}$

$\cos \left(A\right) = \frac{- 84}{-} 96$

$\cos \left(A\right) = \frac{7}{8}$

The second form:

${b}^{2} = {a}^{2} + {c}^{2} - 2 \left(a\right) \left(c\right) \cos \left(B\right)$

Solve for cos(B):

$\cos \left(B\right) = \frac{{b}^{2} - {a}^{2} - {c}^{2}}{- 2 a c}$

Substitute, 4 for a, 6 for b, and 8 for c:

$\cos \left(B\right) = \frac{{6}^{2} - {4}^{2} - {8}^{2}}{- 2 \left(4\right) \left(8\right)}$

$\cos \left(B\right) = \frac{- 44}{- 64}$

$\cos \left(B\right) = \frac{11}{16}$

The third form:

${c}^{2} = {a}^{2} + {b}^{2} - 2 \left(a\right) \left(b\right) \cos \left(C\right)$

Solve for cos(C):

$\cos \left(C\right) = \frac{{c}^{2} - {a}^{2} - {b}^{2}}{- 2 a b}$

Substitute, 4 for a, 6 for b, and 8 for c:

$\cos \left(C\right) = \frac{{8}^{2} - {4}^{2} - {6}^{2}}{- 2 \left(4\right) \left(6\right)}$

$\cos \left(C\right) = \frac{12}{- 48}$

$\cos \left(C\right) = - \frac{1}{4}$