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Law of Cosines & Heron’s Formula Objective: Be able to use Law of Cosines & Heron’s formula to solve oblique triangles as well as find their area. TS:

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Presentation on theme: "Law of Cosines & Heron’s Formula Objective: Be able to use Law of Cosines & Heron’s formula to solve oblique triangles as well as find their area. TS:"— Presentation transcript:

1 Law of Cosines & Heron’s Formula Objective: Be able to use Law of Cosines & Heron’s formula to solve oblique triangles as well as find their area. TS: Explicity assess information and draw conclusions Warm-Up: Recall the Law of Cosines and use it to find a, in the triangle with A=107°, b=10 and c=12.

2 Law of Cosines A B C You could say there are 3 different formulas…. a =________________________________ b = _______________________________ c = ________________________________ But it would be better to notice the pattern. The formula is solving for a side using the two other ___________ and the _________ __________ the side you want to solve for. Note: The Law of Cosines is used when you do not have an angle-side pair. Also, if you are given SSS, it is best to start by solving for the largest angle, to avoid having to worry with the ambiguous case, if you decide to switch to using Law of Sines from there.

3 Solve the triangle where a=17.1, b=10.3 and c=21.7

4 Heron’s Area Formula Proven with the help of Law of Cosines, it is another way you could find the area of a triangle. a b c s = “semi-perimeter” = ____________ A =

5 Solve for θ & y given the below figure is a parallelogram. Then find the area θ 120° y 35 25

6 Solve for θ & y given the below figure is a parallelogram. Then find the area θ 120° y 35 25

7 θ 120° y 35 25 Solve for θ & y given the below figure is a parallelogram. Then find the area

8 Determine if Law of Sines or Cosines would be needed to start solving the triangle a)A = 15°, B = 58°, c = 94 b)a = 96, b = 43, A = 105° c)a = 24, b = 16, c = 29 d)a = 15, c = 42, B = 49°

9 NOW YOU TRY: Solve the triangle a = 31, b = 52, c = 28 Find the area of the triangle above. Answers: A = 29.797°, B = 123.533° & C = 26.670° Area = 361.768 in 2


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