Answer:
2 triangles are possible.
Step-by-step explanation:
Given,
In triangle ABC,
∠ B = 45°,
AC = b = 4 unit,
AB = c = 5 unit,
By the sine law,
[tex]\frac{sin B}{b}=\frac{sin C}{c}[/tex]
[tex]\frac{sin 45^{\circ}}{4}=\frac{sin C}{5}[/tex]
[tex]5 \frac{1}{\sqrt{2}} = 4 sin C[/tex]
[tex]\implies sin C = \frac{5}{4\sqrt{2}} =\implies \angle C\approx 62.114^{\circ}\text{ or }\angle C=117.886^{\circ}[/tex]
If ∠C = 62.114°,
∵ ∠A + ∠B + ∠C = 180°,
∠A + 45° + 62.114° = 180°
⇒ ∠A = 72.886°,
If ∠C = 117.886°
⇒ ∠A + 45° + 117.886° = 180°
⇒ ∠A = 17.114°,
Again by the law of sine,
[tex]\frac{sin B}{b}=\frac{sin A}{a}[/tex]
If ∠A = 72.886°,
[tex]\frac{sin 45^{\circ}}{4}=\frac{sin 72.886^{\circ}}{a}[/tex]
[tex]\implies a\approx 5.01\text{ unit}[/tex]
If ∠A = 17.114°,
[tex]\frac{sin 45^{\circ}}{4}=\frac{sin 17.114^{\circ}}{a}[/tex]
[tex]\implies a\approx 1.66\text{ unit}[/tex]
Hence, two triangles are possible.
