(A) Using the Law of Cosines: $\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}$ Substituting the given values: $\cos 60^{\circ} = \frac{3^{2}+c^{2}-4^{2}}{2(3)

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(A) Using the Law of Cosines: $\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}$ Substituting the given values: $\cos 60^{\circ} = \frac{3^{2}+c^{2}-4^{2}}{2(3)(c)}$ $\frac{1}{2} = \frac{9+c^{2}-16}{6c}$ $\frac{1}{2} = \frac{c^{2}-7}{6c}$ $3c = c^{2}-7$ $c^{2}-3c-7 = 0$

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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In $\Delta ABC$ with usual notations $a=4, b=3, \angle A=60^{\circ}$,then $c$ is a root of the equation

MHT CET 2020 All MHT CET

A
$c^{2}-3c-7=0$
B
$c^{2}-3c+7=0$
C
$c^{2}+3c-7=0$
D
$c^{2}+3c+7=0$

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Relation between sides and angles, Solutions of triangles

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In $\Delta ABC$ with usual notations $a=4, b=3, \angle A=60^{\circ}$,then $c$ is a root of the equation

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