In a triangle $ABC$, if $a=3, b=4$ and $\sin A=\frac{3}{4}$, then $\angle CBA = (\text{in } ^{\circ})?$

  • A
    $60$
  • B
    $75$
  • C
    $90$
  • D
    $45$

Explore More

Similar Questions

The area of triangle $ABC$ is $84 \ sq. \ units$. If $AB = 13$ and $AC = 15$,then $BC$ can be ........ $units$.

In a $\triangle ABC$,$A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$,where $r$ is the inradius and $R$ is the circumradius. Then:

In $\Delta ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} =$

If $A = 60^\circ$,$a = 5$,and $b = 4\sqrt{3}$ in $\Delta ABC$,then $B =$

In $\triangle ABC$,with usual notations,$2ac \sin \left(\frac{1}{2}(A-B+C)\right)$ is equal to

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