The Law of Cosines
1 Use the Law of Cosines to Solve Oblique Triangles
-$$\mathrm { a } = 8 , \mathrm { c } = 8 , \mathrm {~B} = 128 ^ { \circ }$$
A) $\mathrm { b } = 14.4 , \mathrm {~A} = 26 ^ { \circ } , \mathrm { C } = 26 ^ { \circ }$
B) $\mathrm { b } = 17.3 , \mathrm {~A} = 28 ^ { \circ } , \mathrm { C } = 24 ^ { \circ }$
C) $\mathrm { b } = 20.2 , \mathrm {~A} = 24 ^ { \circ } , \mathrm { C } = 28 ^ { \circ }$
D) no triangle

Question

Quizplus · Accepted Answer

The Law of Cosines states that in a triangle with sides a, b, and c, and angles A, B, and C opposite those sides, respectively, then:a^2 = b^2 + c^2 - 2bc cos(A)b^2 = a^2 + c^2 - 2ac cos(B)c^2 = a^2 + b^2 - 2ab cos(C)In this case, we have a = 8, c = 8, and B = 128°. We can use the Law of Cosines to find b:b^2 = a^2 + c^2 - 2ac cos(B)b^2 = 8^2 + 8^2 - 2(8)(8) cos(128°)b^2 = 64 + 64 - 128(-0.61566)b^2 = 64 + 64 + 79.2832b^2 = 207.2832b = sqrt(207.2832)b = 14.4We can also use the Law of Cosines to find A and C:A = cos^-1((b^2 + c^2 - a^2) / 2bc)A = cos^-1((14.4^2 + 8^2 - 8^2) / 2(14.4)(8))A = cos^-1(0.45833)A = 26°C = cos^-1((a^2 + b^2 - c^2) / 2ab)C = cos^-1((8^2 + 14.4^2 - 8^2) / 2(8)(14.4))C = cos^-1(0.45833)C = 26°Therefore, the correct answer is A) b = 14.4, A = 26°, C = 26°.

The Law of Cosines 1 Use the Law of Cosines a=8,c=8, B=128∘\mathrm { a } = 8 , \mathrm { c } = 8 , \mathrm {~B} = 128 ^ { \circ }a=8,c=8, B=128∘

The Law of Cosines
1 Use the Law of Cosines $\mathrm { a } = 8 , \mathrm { c } = 8 , \mathrm {~B} = 128 ^ { \circ }$