Plane Trigonometry Question...

Question

A triangular lot has sides of lengths 242ft, 160ft, 184ft. If the cost per square foot per land is appraised at $40, what is the appraised value of the lot?

Answer 1

(1) Use law of cosines to find one of the angles

Law of cosines: a² = b² + c² - 2bcCosA

Angle A is the angle across from side a, angle B is the angle across from side b, and angle C is the angle across from side c.

 

Let side a = 242, side b = 184, and side c = 160.

a² = b² + c² - 2bcCosA

242² = 184² + 160² - (2)(184)(160)(CosA)

58564 = 33856 + 25600 - (58880)(CosA)

-892 = -58880(cosA)

CosA = 0.01514 [take inverse cos aka cos^-1]

Angle A = 89.13 degrees

 

(2)Calculate the area of the triangle

Use this area formula: Area = (1/2)(b)(c)(SinA)

Area = (0.5)(160)(184)(Sin89.13)

Area = (80)(184)(0.999885)

Area = 14718.31 square feet

 

(3) Multiple the square footage of the triangular land (its area) by the price per square foot to find the lot price:

Price of land = (14718.31 square feet)(40 dollars/1 square foot) = $588,732.43

 

Answer 2

We can use Heron's formula to answer this question.

 

The formula is the square root of the semiperimeter [ given by s = (side a + side b + side c)/2 ] minus a, then the semiperimeter minus b, and then the semiperimeter minus c.

 

Algebraically, Area = √s(s - a)(s - b)(s - c)

 

The value for s = (242f + 160f + 182f)/2 = 293f 

 

In this case, Area = √293f(51f)(133f)(109f)

Area = √216,628,671f⁴

Area  = 14,718f²

 

We then multiply the Area by the cost/f² = 14,718f² x $ = $588,720