(D) Given $3 \cos C = 2$,so $\cos C = \frac{2}{3}$.Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\fr

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(D) Given $3 \cos C = 2$,so $\cos C = \frac{2}{3}$.Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\frac{2}{3} = \frac{9^2 + 8^2 - x^2}{2 \times 9 \times 8}$.$\frac{2}{3} = \frac{81 + 64 - x^2}{144}$.$144 \times \frac{2}{3} = 145 - x^2$.$48 \times 2 = 145 - x^2$.$96 = 145 - x^2$.$x^2 = 145 - 96 = 49$.$x = 7$.

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

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If $a = 9, b = 8$ and $c = x$ satisfies $3 \cos C = 2$,then

A
$x = 5$
B
$x = 6$
C
$x = 4$
D
$x = 7$

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Relation between sides and angles, Solutions of triangles

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In $\Delta ABC$,if $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$,then the values of $a$ and $b$ are:

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If $a = 9, b = 8$ and $c = x$ satisfies $3 \cos C = 2$,then

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The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then the sides of the triangle are

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