Respuesta :
Answer:
The possible solutions for the triangle ABC are:
([tex]a = 5\,cm[/tex], [tex]b = 64.952\,cm[/tex] and [tex]c = 60\,cm[/tex]):
[tex]A = 0.634^{\circ}[/tex], [tex]B = 171.731^{\circ}[/tex]
([tex]a = 5\,cm[/tex], [tex]b = 55.057\,cm[/tex] and [tex]c = 60\,cm[/tex])
[tex]A = 0.751^{\circ}[/tex], [tex]B = 8.293^{\circ}[/tex]
Step-by-step explanation:
The area of the triangle ABC is determined by the following equation:
[tex]A = \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}[/tex]
[tex]s = \frac{a+b+c}{2}[/tex]
Where:
[tex]A[/tex] - Area of the triangle, measured in square centimeters.
[tex]s[/tex] - Semiperimeter of the triangle, measured in centimeters.
[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Sides of the triangle, measured in centimeters.
Now, we simplify the equation:
[tex]A^{2} = s\cdot (s-a)\cdot (s-b)\cdot (s-c)[/tex]
[tex]A^{2} = \left(\frac{a+b+c}{2}\right)\cdot \left(\frac{b+c-a}{2} \right)\cdot \left(\frac{a+c-b}{2} \right)\cdot \left(\frac{a+b-c}{2} \right)[/tex]
[tex]16\cdot A^{2} = (a+b+c)\cdot (b+c-a)\cdot (a+c-b)\cdot (a+b-c)[/tex]
If we know that [tex]A = 25\cdot \frac{\sqrt{3}}{2}\,cm^{2}[/tex], [tex]a = 5\,cm[/tex] and [tex]c = 60\,cm[/tex], the equation is simplified:
[tex]7500 = (b+65\,cm)\cdot (b+55\,cm)\cdot (65\,cm-b)\cdot (b -55\,cm)[/tex]
[tex]7500\,cm^{2} = -(b^{2}-4225\,cm^{2})\cdot (b^{2}-3025\,cm^{2})[/tex]
[tex]7500\,cm^{2} = -(b^{4}-7250\cdot b^{2} +12780625\,cm^{2})[/tex]
[tex]7500\,cm^{2} = -b^{4}+7250\cdot b^{2} -12780625\,cm^{2}[/tex]
[tex]b^{4}-7250\cdot b^{2}+12788125\cm^{2} = 0[/tex]
The roots of the fourth-grade polynomial are:
[tex]b_{1} \approx 64.952\,cm[/tex], [tex]b_{2} \approx 55.057\,cm[/tex], [tex]b_{3}\approx -55.057\,cm[/tex], [tex]b_{4}\approx -64.952\,cm[/tex].
Only the first two roots are physically reasonable. Then, there are only two solutions for the triangle ABC:
[tex]b_{1} \approx 64.952\,cm[/tex], [tex]b_{2} \approx 55.057\,cm[/tex]
The angles A and B can be found by Law of Cosine:
[tex]a^{2} =b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A[/tex]
[tex]2\cdot b\cdot c\cdot \cos A = b^{2}+c^{2}-a^{2}[/tex]
[tex]\cos A = \frac{b^{2}+c^{2}-a^{2}}{2\cdot b\cdot c}[/tex]
[tex]A = \cos^{-1}\left(\frac{b^{2}+c^{2}-a^{2}}{2\cdot b\cdot c} \right)[/tex]
[tex]b^{2} = a^{2}+c^{2}-2\cdot a\cdot c \cdot \cos B[/tex]
[tex]2\cdot a \cdot c\cdot \cos B = a^{2}+c^{2}-b^{2}[/tex]
[tex]\cos B = \frac{a^{2}+c^{2}-b^{2}}{2\cdot a\cdot c}[/tex]
[tex]B =\cos^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2\cdot a \cdot c} \right)[/tex]
Where:
[tex]A[/tex], [tex]B[/tex] - Angles opossed to sides [tex]a[/tex] and [tex]b[/tex], measured in sexagesimal degrees.
There are two possibilities:
([tex]a = 5\,cm[/tex], [tex]b = 64.952\,cm[/tex] and [tex]c = 60\,cm[/tex])
[tex]A = \cos^{-1}\left[\frac{(64.952\,cm)^{2}+(60\,cm)^{2}-(5\,cm)^{2}}{2\cdot (64.952\,cm)\cdot (60\,cm)} \right][/tex]
[tex]A = 0.634^{\circ}[/tex]
[tex]B = \cos^{-1}\left[\frac{(5\,cm)^{2}+(60\,cm)^{2}-(64.952\,cm)^{2}}{2\cdot (5\,cm)\cdot (60\,cm)} \right][/tex]
[tex]B = 171.731^{\circ}[/tex]
([tex]a = 5\,cm[/tex], [tex]b = 55.057\,cm[/tex] and [tex]c = 60\,cm[/tex])
[tex]A = \cos^{-1}\left[\frac{(55.057\,cm)^{2}+(60\,cm)^{2}-(5\,cm)^{2}}{2\cdot (55.057\,cm)\cdot (60\,cm)} \right][/tex]
[tex]A = 0.751^{\circ}[/tex]
[tex]B = \cos^{-1}\left[\frac{(5\,cm)^{2}+(60\,cm)^{2}-(55.057\,cm)^{2}}{2\cdot (5\,cm)\cdot (60\,cm)} \right][/tex]
[tex]B = 8.293^{\circ}[/tex]
The possible solutions for the triangle ABC are:
([tex]a = 5\,cm[/tex], [tex]b = 64.952\,cm[/tex] and [tex]c = 60\,cm[/tex]):
[tex]A = 0.634^{\circ}[/tex], [tex]B = 171.731^{\circ}[/tex]
([tex]a = 5\,cm[/tex], [tex]b = 55.057\,cm[/tex] and [tex]c = 60\,cm[/tex])
[tex]A = 0.751^{\circ}[/tex], [tex]B = 8.293^{\circ}[/tex]
