Algebra 2 problem?

Two sides of a triangular garden have 6 feet and 9 feet of edging. To the nearest tenth, what is the area of the garden if the angle formed by the edged sides is 80 degrees?

Comments

  • Let:

    A= 80 degrees

    a= ?

    b=6 ft

    c= 9 ft

    Solve for side "a" using Cosine Law:

    a^2 = b^2 + c^2 -2bcCosA

    a= sqrt[6^2 + 9^2 - 2(6)(9)Cos80]

    a = sqrt(117 - 18.754)

    a= 9.91 ft

    You can use Hero's formula to directly get the area of the triangle:

    K = sqrt[ s(s-a)(s-b)(s-c)] , where: s=( a+b+c)/2 =(9.91+6+9)/2 = 12.46

    K= sqrt[ 12.46(12.46 - 9.91)(12.46 - 6)(12.46 - 9)]

    K= sqrt[ 12.46(2.55)(6.46)(3.46)]

    K= 26.65 ft^2 <===Answer

    =============================================

    If you are not familiar with Hero's formula, use the Law of sine to solve for one of the two unknown angles, then determine the height of the triangle, and solve for the area.

    a/sinA = b/sinB <===Law of Sine

    9.91/sin80 = 6/sinB

    10.06 = 6/sinB

    sinB= 6/10.06

    B= arcsin(6/10.06)

    B= 36.61 degrees (approx)

    Determine the area of the triangle: (using side "a" as the base)

    A= 1/2(base)(height) ==> sin36.61= h/9 => h= 9sin36.61

    A= 1/2(9.91)(9sin36.61)

    A = 26.60 ft^2 <===Answer

    (Although, there is a decimal difference as compared to Hero's formula.)

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