Question

A triangular 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 measure of the largest angle formed by these streets. Which of the following equations can be used to solve the problem? 2502 = 3002 + 4202 - (2)(300)(420)cosx 3002 = 2502 + 4202 - (2)(250)(420)cosx 4202 = 3002 + 2502 - (2)(300)(250)cos

Answer 1

420^2 = 300^2 + 250^2 - 2(300)(250)cosC

Explanation:

a = 300

b= 250

c = 420

C= ??

Using Cosine rule

Cos C = (a^2 + b^2 - c^2) / 2ab

Cos C = (300^2 + 250^2 - 420^2) / 2*300*250

= (90000 + 62500 - 176400) / 150000

= -23900/150000

= -0.159

C = cos^-1(-0.159)

C = 80.85°

To find angle A

Cos A = (b^2 + c^2 - a^2) / 2bc

Cos A = (250^2 + 420^2 - 300^2) / 2*250*420

= (62500 + 176400 - 90000) / 210000

= 0.71

A = cos^-1(0.71)

A = 44.77°

A+B+C = 180°

B = 180 - (A+C)

B = 180 + (80.85 + 44.77)

B = 54.38°