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. Find the approximate measure of the largest angle formed by these streets.

Answer 1

Consider using the Law of Cosines, because the lengths of three sides are given and the largest angle is the one to be approximated. This angle will be opposite the longest side, that is, opposite the 420-foot side.

420^2 = 250^2 + 300^2 - 2(250)(300)cos A.

Then: 176400 = 62500 + 810000 - 150000cos A.

Solving for cos A, we get:

150000cos A = 176400-62500-810000, or -696100

Then:

-696100

cos A = ------------------- = - 4.64. This is not possible, as the range of the cosine

150000 function is [-1,1].

Answer 2

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, $cosC=\frac{a^2+b^2-c^2}{2ab}$

$cosA=\frac{b^2+c^2-a^2}{2bc}$

and $cosB=\frac{a^2+c^2-b^2}{2ac}$

For finding angle C, we have

$cosC=\frac{300^2+250^2-420^2}{2(300)(25)}$

$cosC=\frac{90,000+62,500-176,400}{150,000}$

$CosC=\frac{-23,900}{150,000}$

$C=cos^{-1}(0.159)$

$A=80.85^{\circ}$ (angle C is in quadrant II)

For finding angle A, we have

$CosA=\frac{250^2+420^2-300^2}{2(250)(420)}$

$CosA=\frac{62,500+176,400-90,000}{210,000}$

$CosA=-0.71$

$A=cos^{-1}(0.71)$

$A=44.765^{\circ}$

For finding angle B, we follow the same procedure, thus

$B=180-(80.85+44.765)= 54.385^{\circ}$

Now, the largest angle is made by angle C, thus third street has the largest angle.