Answer:
There are two unique triangles that can be formed two side lengths of 10 centimeters and one 40º angle.
Step-by-step explanation:
There are theoretically three unique triangles that can be formed with two side lengths of 10 centimeters and one 40º angle. We include a representation of possible triangles:
Case A - Law of Cosine ([tex]a = 10\,cm[/tex], [tex]b = 10\,cm[/tex], [tex]C = 40^{\circ}[/tex])
[tex]c= \sqrt{a^{2}+b^{2}-2\cdot a\cdot b\cdot \cos C}[/tex] (1)
[tex]c = \sqrt{(10\,cm)^{2}+(10\,cm)^{2}-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 40^{\circ}}[/tex]
[tex]c\approx 6.840\,cm[/tex]
Case B - Laws of Sine and Cosine ([tex]a = 10\,cm[/tex], [tex]b = 10\,cm[/tex], [tex]B = 40^{\circ}[/tex])
[tex]\frac{a}{\sin A} = \frac{b}{\sin B}[/tex] (2)
[tex]\sin A = \frac{a\cdot \sin B}{b}[/tex]
[tex]A = \sin^{-1}\left(\frac{a\cdot \sin B}{b} \right)[/tex]
[tex]A = \sin^{-1}\left[\frac{(10\,cm)\cdot (\sin 40^{\circ})}{10\,cm} \right][/tex]
[tex]A = 40^{\circ}[/tex]
[tex]C = 180^{\circ}-A-B[/tex]
[tex]C = 100^{\circ}[/tex]
[tex]c= \sqrt{a^{2}+b^{2}-2\cdot a\cdot b\cdot \cos C}[/tex]
[tex]c = \sqrt{(10\,cm)^{2}+(10\,cm)^{2}-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 100^{\circ}}[/tex]
[tex]c \approx 15.321\,cm[/tex]
Case C - Laws of Sine and Cosine ([tex]a = 10\,cm[/tex], [tex]b = 10\,cm[/tex], [tex]A = 40^{\circ}[/tex])
[tex]\frac{a}{\sin A} = \frac{b}{\sin B}[/tex]
[tex]\sin B = \frac{b\cdot \sin A}{a}[/tex]
[tex]B = \sin^{-1}\left(\frac{b\cdot \sin A}{a} \right)[/tex]
[tex]B = 40^{\circ}[/tex]
[tex]C = 180^{\circ}-A-B[/tex]
[tex]C = 100^{\circ}[/tex]
[tex]c= \sqrt{a^{2}+b^{2}-2\cdot a\cdot b\cdot \cos C}[/tex]
[tex]c = \sqrt{(10\,cm)^{2}+(10\,cm)^{2}-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 100^{\circ}}[/tex]
[tex]c \approx 15.321\,cm[/tex]
There are two unique triangles that can be formed two side lengths of 10 centimeters and one 40º angle.
