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:

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

Explore More

Similar Questions

In triangle $ABC$,$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C = $

With usual notations,in any $\triangle ABC$,if $a \cos B = b \cos A$,then the triangle is:

In $\Delta ABC,$ if ${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3},$ then the triangle is:

The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

Let $ABC$ be an acute-angled triangle with area $R$. Then,$\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2} = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo