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

In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

In a $\triangle ABC$,$\frac{a-b}{a+b} = $

In a $\triangle ABC$,if $\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$,then $B$ is equal to

If $a=3, b=5, c=7$ are the sides of a triangle $ABC$,then $\cot A+\cot B+\cot C=$

In a triangle $ABC$,if $\frac{a}{\tan A}=\frac{b}{\tan B}=\frac{c}{\tan C}$,then $\cos^2 A + \cos^2 B + \cos^2 C =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module