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Unformatted text preview: sin(γ ). After some rearrangement of the
last equation, we get sin(α) = sin(γ ) . If we drop an altitude from vertex A, we can proceed as above
using the triangles ABQ and ACQ to get sin(β ) = sin(γ ) , completing the proof for this case.
B c β c a
γ α h h γ α β c a Q
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 ) .
A B B
c β Q a
α c a β
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,
we have h = c sin(α). Proceeding to B CQ, we get sin(γ ) = h so h = a sin(γ ). Putting this
together with the previous equation, we get sin(γ ) = sin(α) , and we are ﬁnished with this case.
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