Just as you saw used to create an equation for c2, use to create an equation for a2. Solve this equation for h2. ✔ h2 = a2 - x2X h2 = a2 + x2x2 = a2 + h2

Replace your expression for h2 in c2=h2+b2-2bx+x2.✔ C2 = a2 - x2 + b2 - 2bx + x2X c2 = a2 + x2 + b2 - 2bx + x2c2 = -a2 - x2 + b2 - 2bx + x2
Which of the following did you do? Check all of the boxes that apply.
After starting with a non-right triangle, you had to construct the base✔ altitudebisectorhypotenuse.
Use right triangle CDB to create a trigonometric ratio for sin C.



After solving the two sine equations for h, you get h = c · sin A and a = c · sin CX c = h · sin A and a = h · sin C✔ h = c · sin A and h = a · sin C.

What type of triangles did you have to construct to start the proof?
If all of the side lengths of a triangle had a length of 1, what would the simplified law of sines equation be?sin A + sin B + sin C = 1✔ sin A = sin B = sin CX sin A + 1 = sin B + sin C
After forming two right triangles, which right triangle theorem did you have to use twice to solve for side lengths that were not in the final equation?
What does the common side length (h) in each of the sine equations represent in ?ABC?
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