Answer:
Hence, only one triangle is possible.
Step-by-step explanation:
Let, ABC is a triangle.
Given that,
first side of a ΔABC is 4, second side of a ΔABC is 3 and ∠B is [tex]41[/tex]°.
Diagram of the ΔABC is shown below.
Now,
[tex]\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}[/tex] (Law of sines)
[tex]\frac{sinA}{a}=\frac{sinB}{b}[/tex]
⇒ [tex]\frac{sinA}{4}= \frac{sin41}{3}[/tex]
⇒ [tex]sinA= \frac{sin41\times 4}{3}[/tex] =[tex]0.87475[/tex]
⇒ [tex]sin^{-1} (0.87475)=A[/tex]
⇒ ∠A = 61°
Now, ∠A + ∠B+ ∠C = 180° (angle sum property)
⇒ 61° + 41° + ∠C = 180°
⇒ ∠C = 78°
[tex]\frac{sinB}{b}=\frac{sinC}{c}[/tex]
⇒ [tex]\frac{sin41}{3}=\frac{sin78}{c}[/tex]
⇒ [tex]c=\frac{sin78\times3}{sin41}[/tex]
⇒ [tex]c= \frac{0.9782\times3}{0.65606} =4.47[/tex]
Therefore, ∠A = 61°, ∠B = 41°, ∠C = 78°, [tex]a=4, b=3, c= 4.47.[/tex]
Hence, only one triangle is possible.
