If a = 2 sqrt{2}, b = 6, and A = 45^{circ}, t | vedclass

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

Question

(A) Given: $a = 2 \sqrt{2}$, $b = 6$, and $A = 45^{\circ}$.Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the values: $\fr

Vedclass · Accepted Answer

no triangle is possible

Vedclass · Answer

(A) Given: $a = 2 \sqrt{2}$, $b = 6$, and $A = 45^{\circ}$.Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the values: $\frac{2 \sqrt{2}}{\sin 45^{\circ}} = \frac{6}{\sin B}$.$\sin B = \frac{b \sin A}{a} = \frac{6 	imes \sin 45^{\circ}}{2 \sqrt{2}}$.$\sin B = \frac{6 	imes \frac{1}{\sqrt{2}}}{2 \sqrt{2}} = \frac{6}{2 	imes 2} = \frac{6}{4} = 1.5$.Since the value of $\sin B$ cannot exceed $1$, $\sin B = 1.5$ is impossible.Therefore, no triangle is possible.

If $a = 2 \sqrt{2}$, $b = 6$, and $A = 45^{\circ}$, then:

Similar Questions

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

In $\triangle ABC$, if $\angle C = \frac{\pi}{3}$, then the value of $\cos^2 A + \cos^2 B - \cos A \cos B$ is...

In a $\Delta ABC$,the ratio of angles $A : B : C = 3 : 5 : 4$. Then $a + b + c \sqrt{2}$ is equal to

In a triangle $ABC$,if $a=3, b=7, c=8$,then $\sin \frac{B}{2} \tan \frac{C-A}{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module