(B) Using the sine rule,we have: $\frac{\sin A}{a} = \frac{\sin B}{b}$ Substituting the given values: $\frac{\sin 60^{\circ}}{6} = \frac{\sin B}{8}$ S

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(B) Using the sine rule,we have: $\frac{\sin A}{a} = \frac{\sin B}{b}$ Substituting the given values: $\frac{\sin 60^{\circ}}{6} = \frac{\sin B}{8}$ Since $B = \sin^{-1} x$,we have $\sin B = x$. $\frac{\sqrt{3}/2}{6} = \frac{x}{8}$ $\frac{\sqrt{3}}{12} = \frac{x}{8}$ $x = \frac{8\sqrt{3}}{12} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$

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

Medium

In a triangle $ABC$,with usual notations,if $m \angle A = 60^{\circ}$,$b = 8$,$a = 6$,and $B = \sin^{-1} x$,then $x$ has the value:

MHT CET 2023 All MHT CET

A
$\frac{\sqrt{3}}{2}$
B
$\frac{2}{\sqrt{3}}$
C
$2\sqrt{3}$
D
$\frac{1}{2\sqrt{3}}$

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Class 11

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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 $ABC$,with usual notations,if $m \angle A = 60^{\circ}$,$b = 8$,$a = 6$,and $B = \sin^{-1} x$,then $x$ has the value:

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