(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Given $a = 4$,$b = 3$,and $A = 60^\circ$,we have: $\cos 60^\circ = \frac{3^2 + c^

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(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Given $a = 4$,$b = 3$,and $A = 60^\circ$,we have: $\cos 60^\circ = \frac{3^2 + c^2 - 4^2}{2(3)(c)}$ $\frac{1}{2} = \frac{9 + c^2 - 16}{6c}$ $\frac{1}{2} = \frac{c^2 - 7}{6c}$ $6c = 2(c^2 - 7)$ $6c = 2c^2 - 14$ $2c^2 - 6c - 14 = 0$ Dividing by $2$,we get: $c^2 - 3c - 7 = 0$.

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

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In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

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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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

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