# 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 Relevance
• 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

• 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

• 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

• 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

:)>