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

If the area of triangle $ABC$ is $b^2-(c-a)^2$,then $\tan B=$

In a $\triangle ABC$,if $a^4+b^4+c^4=2b^2c^2+2a^2b^2$,then $B=$

If $b = 3, c = 4$ and $B = \frac{\pi}{3}$,then the number of triangles that can be constructed is

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

In $\Delta ABC$,$8 \Delta = (b + c)(bc + 1)$,then the circumradius of $\Delta ABC$ is (where $\Delta$ denotes the area of the triangle and $b, c$ are the lengths of sides $AC$ and $AB$ respectively):

Vedclass Test Series

Exam Paper Generator

Online Exam Module