If A = 60^circ,a = 5,and b = 4sqrt{3} in Delt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Rearranging for $\sin B$: $\sin B = \frac{b \sin A}{a}$.Substituting the given value

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Rearranging for $\sin B$: $\sin B = \frac{b \sin A}{a}$.Substituting the given values: $\sin B = \frac{4\sqrt{3} \sin 60^\circ}{5} = \frac{4\sqrt{3} 	imes (\frac{\sqrt{3}}{2})}{5} = \frac{4 	imes 3}{2 	imes 5} = \frac{6}{5} = 1.2$.Since the value of $\sin B$ must be $\le 1$,the value $1.2$ is inadmissible.Therefore,no such triangle exists,and the correct option is $D$.

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

Similar Questions

In $\Delta ABC$,$\frac{\sin(A - B)}{\sin(A + B)} = $

In the triangle $ABC$,if $a=7, b=6$ and $A=120^{\circ}$,then the approximate value of $B$ is (in $^{\circ}$)

In $\Delta ABC,$ $a\sin (B - C) + b\sin (C - A) + c\sin (A - B) = $

If in a triangle $ABC$,$a = 5$,$b = 4$,and $A = \frac{\pi}{2} + B$,then $C$ is:

In triangle $ABC$,if $r_1=36, r_2=18$ and $r_3=12$,then $a+b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module