Answer:
Step-by-step explanation:
According to the question,A triangular city lot bounded by three streets has a length of 300 feet on one street, 250 feet on the second, and 420 feet on the third, thus
Let the angles be A, B, and C and the sides a, b, c.
a=300, b=250 and c=420
Then, [tex]cosC=\frac{a^2+b^2-c^2}{2ab}[/tex]
[tex]cosA=\frac{b^2+c^2-a^2}{2bc}[/tex]
and [tex]cosB=\frac{a^2+c^2-b^2}{2ac}[/tex]
For finding angle C, we have
[tex]cosC=\frac{300^2+250^2-420^2}{2(300)(25)}[/tex]
[tex]cosC=\frac{90,000+62,500-176,400}{150,000}[/tex]
[tex]CosC=\frac{-23,900}{150,000}[/tex]
[tex]C=cos^{-1}(0.159)[/tex]
[tex]A=80.85^{\circ}[/tex] (angle C is in quadrant II)
For finding angle A, we have
[tex]CosA=\frac{250^2+420^2-300^2}{2(250)(420)}[/tex]
[tex]CosA=\frac{62,500+176,400-90,000}{210,000}[/tex]
[tex]CosA=-0.71[/tex]
[tex]A=cos^{-1}(0.71)[/tex]
[tex]A=44.765^{\circ}[/tex]
For finding angle B, we follow the same procedure, thus
[tex]B=180-(80.85+44.765)= 54.385^{\circ}[/tex]
Now, the largest angle is made by angle C, thus third street has the largest angle.
