# Answer to Question #14838 in Trigonometry for Nathan

Question #14838

A triangular piece of land is bounded by three straight roads, two of which intersect in a right angle. The third road makes an angle of 37.6° with one of the others. The perimeter of the land is 985 yards. What is its area, to the nearest square foot?

Expert's answer

Let's denote the road triangle's sides by a, b, and c. Then we've got the following system of equations:

a + b + c = 985;

a = c* cos(37.6);

b = c* sin(37.6);

Let's plug in last two into first to solve for c:

c*cos(37.6) + c*sin(37.6) + c = 985 ==>

0.79c + 0.61c + c = 985 ==>

2.4c = 985 ==>

c = 410

now plug c into those initial equations for a and b to solve for a and b:

a = 326, b = 250: area = 1/2 * a*b = 0.5*326*250 = 40.750 square yards.

a + b + c = 985;

a = c* cos(37.6);

b = c* sin(37.6);

Let's plug in last two into first to solve for c:

c*cos(37.6) + c*sin(37.6) + c = 985 ==>

0.79c + 0.61c + c = 985 ==>

2.4c = 985 ==>

c = 410

now plug c into those initial equations for a and b to solve for a and b:

a = 326, b = 250: area = 1/2 * a*b = 0.5*326*250 = 40.750 square yards.

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