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

Answer 1

Answer:

The approximate measure of the largest angle formed by these streets is $99.2\°$

Step-by-step explanation:

we know that

Applying the law of cosines

$c^{2}=a^{2}+b^{2} -2(a)(b)cos(C)$

In this problem we have

$a=300\ ft$

$b=250\ ft$

$c=420\ ft$ ----> is the greater side

substitute and solve for angle C

$420^{2}=300^{2}+250^{2} -2(300)(250)cos(C)$

$176,400=152,500 -150,000cos(C)$

$cos(C)=[152,500-176,400]/150,000$

$cos(C)=-0.1593$

$C=arccos(-0.1593)=99.2\°$