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Solve the triangle. A = 54°, b = 11, c = 8 No triangles possible a ≈ 13.4, C ≈ 49.6, B ≈ 76.4 a ≈ 13.4, C ≈ 45.6, B ≈ 80.4 a ≈ 9, C ≈ 45.6, B ≈ 80.4
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Use cosine law for angle A : \[\large{\cos A = \cfrac{b^2 + c^2 - a^2}{2bc} } \]
\[\large{\cos 54 = \cfrac{11^2 + 8^2 - a^2}{2(11)(8)} }\] \[\large{0.58 = \cfrac{121 + 64 - a^2 }{176} \\ 0.58 = \cfrac{185 - a^2 }{ 176 } \\ (0.58)(176) - 185 = -a^2 \\ 102.08 - 185 = -a^2 \\ -82.92 = -a^2 \\ 82.92 = a^2 \\ a = \sqrt{82.92} = 9 (approx.) \\ }\] it is clear from the options that only the third option is having a as 9 units, so it is c)
@tholn, hope it helped.
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