Precalculus/trigonometry find area of triangle?

Find the area of each triangle.

a. AB = 7 cm, AC = 5 cm, angle A = 27 degrees

b. BC = 4 ft, AB = 7 ft, AC = 5 ft

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4 Answers

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  • 1 month ago

    a) base * height / 2

    7*5*sin(27) /2 = 7.94cm²

    b) use law of cosine to discover a degree. then base * height /2

    a²+b²-2abCOS(°) = c²

    a²+b²-c²=2abCOS(°)

    5²+7²-4²= 2abCOS(°) = 58

    arccos(58/2/5/7) = 34.05°

    7*5*sin(34.05) /2 = 9.8ft

  • 1 month ago

    Find the area of each triangle. 

    a. 

    AB = 7 cm, AC = 5 cm, angle A = 27 degrees

    Area = 7.94483 cm^2

    b. 

    BC = 4 ft, AB = 7 ft, AC = 5 ft 

    Area = 9.79796 ft^2

  • 1 month ago

    a)

    Area of triangle ABC = (1/2) AB*AC*sin27° 

    = (1/2) 7*5*sin27° 

    = 7.945 cm²

    ≈ 7.9 cm²

    b)

    Area of triangle ABC = (1/4) √((AB+BC+AC)(AB+BC-AC)(AB-BC+AC)(-AB+BC+AC)) 

    = (1/4) √((7+4+5)(7+4-5)(7-4+5)(-7+4+5))

    = (1/4) √(16*6*8*2)

    = 9.7979589... ft²

    ≈ 9.8 ft²

  • 1 month ago

    Area is (1/2)absinC....where a and b are adjacent sides and C is the angle between them.

    a) (1/2) x 7 x 5 x sin27 => 7.9 cm²

    b) We need to use the 'cosine rule' to find an angle.

    so, a² = b² + c² - 2bcCosA

    Then, 4² = 5² + 7² - 2(5)(7)CosA

    => CosA = (5² + 7² - 4²)/70

    i.e. CosA = 58/70

    Hence, A = 34° 

    Using (1/2)absinC again we have:

    (1/2) x 5 x 7 x sin34° => 9.8 ft²

    :)>

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