115 graphs of polar equations 809 coordinate system

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Unformatted text preview: sin(γ ). After some rearrangement of the c a last equation, we get sin(α) = sin(γ ) . If we drop an altitude from vertex A, we can proceed as above a c using the triangles ABQ and ACQ to get sin(β ) = sin(γ ) , completing the proof for this case. b c B B B c β c a γ α h h γ α β c a Q γ CA CA C Q b b For our next case consider the triangle ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sin(γ ) and h = c sin(β ) so that sin(β ) = sin(γ ) . b c A B B c β Q a α c a β h γ A b γ C A b C Dropping an altitude from vertex B also generates two right triangles, ABQ and B CQ. We know that sin(α ) = h so that h = c sin(α ). Since α = 180◦ − α, sin(α ) = sin(α), so in fact, c we have h = c sin(α). Proceeding to B CQ, we get sin(γ ) = h so h = a sin(γ ). Putting this a together with the previous equation, we get sin(γ ) = sin(α) , and we are finished with this case. c a 3 4 Your Science teache...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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